With 22 practice questions, I'm going to teach you the entire grade 10 math course in just one hour. The Ministry of Education says it should take 110 hours to learn all of the material. Let me see if I can save you some time. This video will be a great way to review the material you have already learned, or to get introduced to the types of questions taught at this grade level.
Here we go. Example one, solve the following linear system in three ways by graphing it, substitution, and elimination. With all three methods we should get the same answer, but we'll go through how you do each of them. Now when solving a linear system, basically what that means is we have two linear equations, we want to figure out what's the values of the variables that make both equations true. We can do that algebraically with substitution or elimination, or graphically by graphing both lines and seeing where the point of intersection is.
So let's do graphing first. In order to graph both of these lines, it's easiest if we rearrange them into y equals mx plus b form. So I'm going to isolate y in both of these equations. Let me start with the x minus y equals 5 equation.
Let me isolate y. It would be easiest if I isolate it to the right. And now you can see it's in y equals mx plus b form, where m is the slope, and in this case our m value is 1, and b is our y-intercept, and in this case our b value.
is negative 5. So we could use that slope and y-intercept to graph the line. So we would plot the y-intercept of negative 5, and then use the slope of 1. Remember, slope is rise over run, so you may want to rewrite that whole number 1 as a fraction 1 over 1. So rise 1, run 1, plot a point, and keep plotting points until you fill the grid. And then we'll connect them with a straight line, and we'll go through the same process for the other line.
Let me isolate y, I'll just move the 3x term to the right, becomes negative 3x. So it's clear that for this one, the slope is negative 3, and the y-intercept is 3. Plot the y-intercept, and then use the slope. And remember, any whole number is over 1, so you can think of this negative 3 as negative 3 over 1. So rise negative 3, run 1. That means down 3, right 1. And keep plotting points until you fill your grid.
Connect them with a straight line. and then just find where these lines intersect and that's the solution to the system. These lines intersect right at that point right there, the point.
That's our point of intersection. So we can say the solution is, you could write it one of two ways, you could write it as a point, the point, or you could write what x and y are equal to. You could write x equals 2, y equals negative 3. Either would be acceptable ways of writing the final answer.
Let's solve that same system. using elimination and substitution. Elimination first. So in doing elimination you want the equations in the format x plus y equals the constant, which both equations are already in that format.
Now our goal here is to eliminate a variable, and the way we do that is by making the coefficients of either the x's or the y's have the same absolute value. Now in this example the coefficients of the y's already have the same absolute value. The coefficient of y in the first equation is negative one. and the coefficient of y in the second equation is positive one.
Those have the same absolute value, but because they're opposite signs, adding the equations will make them eliminate. If they were the same sign, subtracting them would make them eliminate. But in this case, since they're the same absolute value but opposite signs, if we add these two equations together, the y's will eliminate. Watch. Let's add the equations, and when we add the equations we make sure we collect the like terms.
That's why we have them lined up like this. x plus 3x is 4x. Negative y plus y is 0, that they eliminate, which is what we want, and 5 plus 3 is 8. So we have 4x equals 8, divide both sides by 4, we get x equals 2. So we have the value of x that make both equations the same, but what's the value of y?
We now need to sub x equals 2 into either of the original equations. It doesn't matter which one you plug it into, because that's the value of x that makes both equations have the same value of y. I'll choose the first one, it looks easier.
So sub 2 in for x, 2 minus y equals 5. I'll isolate y, move the negative y to the right, move the 5 to the left, I've got 2 minus 5 equals y, negative 3 equals y. So notice my solution, x equals 2, y equals negative 3. That's the same thing we got by solving it graphically. Let's try it again, but using substitution. Now when doing substitution, I often like to line them up side by side like this.
I'll still number the equations, equation 1 and equation 2. When doing substitution, what you want to do is isolate one of the variables in either of the equations. It doesn't matter. I think in equation 1, the variable x looks like it would be easy to isolate since it already has a coefficient of 1. So I'll pick this x to be the variable that I'm going to isolate. So I'll move this negative y to the right, and it becomes positive y. So I have x equals 5 plus y.
Now what you do is you take... what the variable you isolated is equal to, so it says x equals 5 plus y, you take that and plug it into the other equation for x, because we know x is equal to 5 plus y. So now in the other equation I have 3 times 5 plus y plus y equals 3. Now we have an equation with only one variable, this equation only has the variable y, we could solve this for y.
Let me distribute the 3 into the brackets. Collect my like terms, I've got 15 plus 4y equals 3. Now I want to isolate the y, so I'll move the 15 over. 4y equals 3 minus 15, which is negative 12. Divide both sides by 4, I get y equals negative 3. I now need to solve for the other variable. So what we have to do is take that answer and plug it back into either of the original equations. It's always easiest if you sub it back into this rearranged version we have, because x is isolated, so you just take this answer for y we got.
plug it back in for y there and then we get our answer for x so i'd have x equals 5 plus y which we know is negative 3. so i have x equals 5 plus negative 3 that's 5 minus 3 which is 2. so i got x equals 2 y equals negative 3 same answer again just a different method example 2 solve the following linear system using elimination so this one's much harder than the first one we did for this one we have two equations again and both equations are already in the proper format with x and y on one side of the equation and the constant on the other side, but we want to eliminate the variables by adding or subtracting the equations, and to do that the coefficients have to have the same absolute value. And notice the x's have different coefficients and the y's have different coefficients. So what we're going to have to do is multiply one or both of the equations by a constant to make the coefficients have the same absolute value.
What sticks out to me is we can make the x's have a coefficient of 20, or we can make the y's have a coefficient of 12 and negative 12. Doesn't matter. Let's make the x's both have a coefficient of 20. So I would have to multiply equation 1 by 5. So I'm going to do 5 times equation 1, and that would give me, so 5 times 4x is 20x, 5 times 3y is 15y, and 5 times 13 is 65. Make sure you multiply every term in the equation by the constant. And for the second equation, for the x to have a coefficient of 20, I'd have to multiply it by 4. So 4 times equation 2. and that would give me 20x 4 times negative 4y is negative 16y and 4 times negative 7 is negative 28. now what i'm going to do is i'm going to figure out should i add or subtract these to eliminate the variable the x's have the exact same coefficient with the exact same sign they're both positive 20 so subtracting will eliminate the variable x so let's subtract all the like terms 20x minus 20x is 0x so that's gone it eliminates 15y minus negative 16y is 31y, and 65 minus negative 28 is 93. Divide both sides by 31, and we get y equals 3. Now we need to sub y equals 3 into either original equation. So I'll just sub it into equation 1. Let's sub our answer for y into that equation. So 3 times 3 equals 13, and then we just solve this for x.
So 4x plus 9 equals 13. Let's subtract the 9 to the other side, I get 4x equals 4, therefore x must equal 1. So my solution to this linear system is x equals 1, y equals 3. If we were to graph both lines, they would intersect at the point. Example 3. Here's an application of solving linear systems. A sports shop sells adidas shoes for $82 a pair and air jensen basketball shoes for $95 a pair.
One day the shop sells a combined 75 pairs of shoes. totaling $6,241 in sales. How many pairs of each shoes were sold? Now for these types of questions, they're always going to want you to solve for two variables, and in order to solve for two variables, you must be able to make two equations that involve those two variables. So I always like to start by figuring, okay, what variables am I going to be trying to solve for?
So it wants to know how many pairs of each shoe are sold. So we're trying to figure out, okay, how many Adidas shoes were sold and how many Air Jensen shoes were sold. Those are my variables.
I'll say x equals number of Adidas and y equals number of air jensen. And in order to figure out what the value of these variables are, I must be able to make two equations that involve these two variables. Well I know they sold a combined 75 pairs of shoes.
So my first equation x plus y equals 75. There's one equation that involves x and y. In the second equation, well I know what the total dollar amount in sales was, and I know how much each pair costs. Adidas shoes cost $82 a pair, so 82 times the number of Adidas sold. plus, air jensen's cost $95, so 95 times the number of air jensen's sold would equal that $6,241. when you have two equations with two variables you can solve them by graphing, substitution, or elimination.
i think i'll do elimination for this. let me make the coefficients of the x's be the same, so i'll multiply the first equation by 82, and that would give me 82x plus 82y, and 82 times 75 is 61.50. Now let me rewrite equation two, lined up with that right underneath of it.
And since the coefficients of the x have the exact same sign, I know subtracting will eliminate the variable. 82x minus 82x is 0, 82y minus 95y is negative 13y, and 6150 minus 6241 is negative 91. Now if I divide both sides by negative 13 I get y equals 7. So I know that they sold seven pairs of Air Jensen shoes. Let me just sub that answer back into either of the original equations. Let me sub it back into one that's going to be a lot easier.
and then solve for x. So I'll sub 7 in for y, therefore x is 68. So our final answer for this, they sold 68 pairs of adidas and 7 pairs of air jensen. Next you need to know how to calculate the midpoint and distance of a line that connects two points. So I've got point a and point b, it says calculate the midpoint and the distance between those points.
Let's start with midpoint. If you notice the formula, it says we need the x and y coordinates from our first and our second point. Now it doesn't matter which your first point is and which your second point is, but it's helpful to pick one and label it.
So I'll make a my first point, which means five is my x one and negative three is my y one, it'll make be my second point, which means this is my x two, and this is my y two, right, each point has an x and a y. If I want the midpoint between these two, using the formula, it tells me that what I need to do is average the x coordinates. So I'll add x one and x two together.
So I'll do five plus negative one. And then I have to divide it in two to get the average of the two of them, right, the middle of the two points is going to be at the average of the x coordinates, and the average of the y coordinates. So I now need to average the y coordinates by adding them and dividing by two. So negative three, plus five, divided by two. And the midpoint, if I simplify this, the x coordinate, 5 plus negative 1 over 2, that's 4 over 2, which is 2, and the y coordinate, negative 3 plus 5 is 2, divided by 2 is 1. Let's now find the distance between point A and B.
The distance formula comes from Pythagorean theorem. To find the distance, I do the square root of the difference in the x coordinates, so the formula tells me to do x2 minus x1, and then square it, so negative 1 minus 5 squared, plus the difference in the y coordinates. So now I have to do y2 minus y1, so 5 minus negative 3, and then square that difference.
And I just have to simplify this. So negative 1 minus 5 is negative 6, square it, and I get 36. 5 minus negative 3 is 8, square it, and I get 64. So I need the square root of 36 plus 64. That's the square root of 100, which is 10. So the distance between the two points is 10 units. I don't know what the units are, but that's the distance between the two points.
Example 5 says draw the triangle with these vertices. So let me start by plotting those vertices. So there's my triangle.
It says draw the median from C to AB, then find the equation of this median. Remember, a median of a triangle is the line segment that joins a vertex to the midpoint of the opposite side. So the median from C to AB would be the line that goes from C to the middle of AB.
So let me just estimate where that is right here. It looks like it's going to be about there. That's the middle of AB.
So the median would go from C to that point. So what I've drawn in red is the median. Let's find the equation of that line in the form y equals mx plus b. So I'm going to need to know its slope and y-intercept.
Let's start by figuring out its slope. And to figure out its slope, I would need to know two points on the line. Well, I know it's on point C, and it's also at the midpoint of AB. Okay, so I'm going to need to know the midpoint of AB. Let's find the midpoint.
So the midpoint of AB, I'm going to need to average the x coordinates, so negative 1 plus 1 divided by 2, and average the y coordinates, 3 plus 5 over 2. So the midpoint is going to be negative 1 plus 1 is 0, so 0 over 2 is 0, and 8 over 2 is 4. So my midpoint is 0, 4. And I know point C is on the median as well, which the question tells us is the point 3, 1. I could use those two points to find the slope of that median. I'll call the slope M, and remember to calculate slope, you just do change in y over change in x. y2 minus y1 over x2 minus x1. So I'll make C my second point, and the midpoint of A be my first point.
So 1 minus 4 over 3 minus 0. And that gives me negative 3 over 3, which is negative 1. So the slope of my median is negative 1. If I want the y-intercept of the median, I'm going to have to, into the equation y equals mx plus b, I'm going to have to sub in the slope I know and a point on the line. It doesn't matter which point. I can pick the point 0, 4. And actually, that's already the y-intercept.
I know the y-intercept is 4, but let me just show you algebraically how we would get it. So we would plug in the point 0, 4 for x and y. The y-coordinate's 4. The x-coordinate is 0. And we know m is negative 1. And we solve for b.
4 equals 0 plus b. So b is 4. So now that I have the slope and the y-intercept of the median, I could write the equation. y equals, we sub in 4, m, and b.
So negative 1x plus 4. Which you would just write as negative x plus 4. So that's my final answer. That's the equation of that red line. Example 6. Determine the equation for the right bisector of the line segment with the endpoints A and B.
Let's remember what a right bisector is. A right bisector, often called a perpendicular bisector, is a line that is perpendicular to a line and passes through its midpoint. So I know my perpendicular bisector is going to be perpendicular to the line connecting A and B.
So I'm going to need to know the slope of A and B. which I know I get by doing change in y over change in x, so y2 minus y1, 6 minus negative 4, over x2 minus x1, 8 minus negative 2, so that gives me 10 over 10, which is 1, and I know my right bisector is going to be perpendicular to this line, so it's going to have a slope perpendicular to 1. Perpendicular slopes are what are called negative reciprocals of each other, so I would have to take this 1, which you could rewrite as 1 over 1, flip it, and change its sign. Well the reciprocal of 1 over 1, if we flip that it's still 1 over 1, but we do have to change its sign so it becomes negative 1 over 1, so negative 1. So our slope of our right bisector, I'll call that the perpendicular slope, that equals the negative reciprocal of 1 over 1 which is negative 1 over 1, which is just negative 1. If I want the equation of the right bisector, not only do I need its slope but I need its y intercept, and to find that I'm going to need to know a point on the right bisector.
Well, I know the right bisector goes through the midpoint of AB, so I'm going to need to know the midpoint of AB. So I'll average the x coordinates, negative 2 plus 8 over 2, and average the y coordinates, negative 4 plus 6 over 2. So my midpoint, the x coordinate is 6 over 2, which is 3. The y coordinate would be 2 over 2, which is 1. So my midpoint is 3, 1, which is an x and a y coordinate. I now take that x and y and the slope of my right bisector, and I can solve for b.
the y-intercept. So I'll sub in my point for x and y, and negative 1 for my slope, and if I solve this for b, I get b equals 4. So I can now write my final equation by subbing in my m and my b values into y equals mx plus b. So y equals, and make sure you use your perpendicular slope, because we're doing the right bisector equation, so negative x plus the b value, 4. So there's my final answer. Example 7, classify the triangle with these vertices as either scalene, isosceles, or equilateral, and also state if it has a right angle. Let me start by just drawing it quickly for you.
Alright, there we go. It looks like it's probably scalene. Angle D looks like it might be close to 90 degrees.
We'll have to do some calculations to know for sure. So if I want to classify the triangle, and give proof as to what shape it is, I'm going to need to know the length of each line, which we can use our distance formula for. So let's find the distance of all three side lengths.
So let me start with side DE. Now you can do DE or ED, doesn't matter. I'll do DE.
If I'm doing DE, I'll make D my first point, X1, Y1, and E my second point, X2, Y2. Plug that into my distance formula and get the length of side DE. So I would have the square root of difference in the X's squared.
So negative two minus negative four squared, plus the difference in the Y squared, six minus negative two squared. If I simplify underneath the square root, I'd have two squared, which is four plus eight squared, which is 64. So four plus 64 is 68. So this would be root 68, we would do the same thing for the other two sides. So for side E F, and D F. And then we'll have all three side lengths.
For time's sake of this video, I'm just going to fast forward through me solving for these side lengths. Okay, I have all three side lengths. Notice all three sides are different lengths, therefore this is a scalene triangle. But does it have a right angle? Well, I know that right angle triangles, Pythagorean theorem is true.
The sums of the squares of the shorter two sides is equal to the square of the longer side. So let's check. Is the square root of 68 squared plus the square root of 104 squared, is that equal to the square of the longest side? So root 164 squared. If this is true, then there's a right angle.
If it's not true, then there's no right angle. Square rooting and squaring are inverse operations. They cancel each other out. So really what we have here is 68 plus 104. Is that equal to 164?
It's not. 172 does not equal 164. Therefore, there's no right angle. Therefore, not a right triangle. Example 8, this is a really quick one. What's the radius of a circle?
Remember the equation of a circle is x squared plus y squared equals r squared. In this case, r squared is 36, so the radius squared is equal to 36, which means the radius would be equal to the square root of that, and the square root of 36 is 6. We don't have to consider the negative square root of it because radius lengths can't be negative, we only look at the positive, so the radius is 6 units. 9, for the circle that's centered at the origin and passes through the point negative 3, 4, So center at the origin passes through this point, negative 3, 4. Let me just draw a circle that does that really quickly for you.
That circle, we want to find the equation of this circle. And part B asks us to check if this point 5, 2 lies on the circle, inside of it, or outside of it. And we'll do this algebraically.
So if I want the equation of this circle, well, I know one point on the circle, this point right here, negative 3, 4. That's an x, y point. The equation of a circle is of the form x squared plus y squared. equals r squared.
So I need to know what's the radius squared to figure that out, just plug in your x and y into the equation. So negative three squared plus four squared is equal to the radius squared, I have nine plus 16 equals r squared. So r squared is 25, which means the radius is five, but I don't need the radius in this question. I just need the radius squared so that I can write the equation because the equation is x squared plus y squared equals r squared.
which is 25. So my final equation would be x squared plus y squared equals 25. That's the equation of this circle. Now if we want to check if this point lies inside the circle, outside of it or on the circle, we just plug five two into the equation of the circle and see if we get a number that's less than 25 equal to 25, or greater than 25. That'll tell us whether it lies inside, outside or on the circle. And we can tell from the graph that the point five two This point right here is clearly outside of the circle, but if we don't have the graph and we want to prove it algebraically, like I said, we take our equation of our circle and we see what happens when we plug this value in.
So if we plug 5 and 2 in for x and y, well 5 squared plus 2 squared, that's actually going to give me a value that's bigger than 25, right? 25 plus 4 is 29. 29 is bigger than 25. That tells me that the point must be outside the circle. If we got 25 equals 25, we'll be on the circle. If we got something less than 25, like we got 21 less than 25, I mean inside the circle.
Example 10 says what's the shortest distance from point negative three five to this line y equals a quarter x plus 10. This question is going to involve pretty much everything from this unit. So we've got the line y equals a quarter x plus 10. I'll just draw an approximation of that line there for you there, just to visualize what's happening here. And we have the point negative three five. I don't have a Cartesian grid, but let me just put it right here. Here's the point negative 3, 5. I'm looking for the shortest distance from that point to that line.
The shortest distance from the point to this line isn't going to be something like that or like that. It's going to be along a line that's perpendicular to the original line. So it's going to be along that line right there, which forms a 90-degree angle with the original line y equals a quarter x plus 10. This black line is y equals a quarter x.
plus 10. And I want to figure out the shortest distance. So basically the distance from this point to whatever this point is here, I don't have that point, but I'm definitely going to need it to find that point. Well, that's where the blue line intersects the black line. So if I have the equation of the blue line, I could figure out where they intersect. So to find the equation of the blue line, well, I need it slope and y intercept.
Well, I know it's perpendicular to y equals a quarter x plus 10. So I know its slope is going to be the negative reciprocal of a quarter. So the perpendicular slope would equal the negative reciprocal of a quarter, which means flip it and change the sign. So it's negative 4 over 1, which is just negative 4. To write the equation of that blue line, I also need the y-intercept, the b value.
To get that into y equals mx plus b, I have to plug in a point on the line and the slope. Well, I know a point, negative 3, 5. It has an x and a y, so sub that in. 5 for y. I know the slope of the blue line is negative 4, I know x is negative 3. If I solve this equation for b, negative 4 times negative 3 is 12, subtracted over 5 minus 12 is negative 7. So my b value is negative 7, so the equation of that blue line is y equals negative 4x minus 7. So this blue line, y equals negative 4x minus 7. Now let's figure out where does that blue line intersect the black line.
I've got two equations I can solve using substitution or elimination. Let me rewrite the two equations. And I think since y is already isolated in both equations, I think substitution is the easiest method.
So really, all we do is we take what y is equal to from one equation and replace the y in the other equation with that. So replace the y in the second equation with a quarter x plus 10. And now let me solve this equation. I don't really like this fraction here.
So what I'm going to do is I'm going to get rid of the fraction by multiplying both sides of the equation by its denominator. So I'll multiply both sides by four. Now give me x. plus 40 equals negative 16 x minus 28. Now I'll collect my like terms onto the same sides of the equation.
I've got 17 x equals negative 68. So I've got x equals negative four, I now need to figure out what y equals. So let me just sub x equals negative four into either original, I'll just plug it into the first one because I have room underneath of there. So I'll plug negative four in for x.
A quarter times negative 4 is negative 1, so I've got negative 1 plus 10, which is 9. So my point of intersection is the point negative 4, 9. So I know this point right here is the point negative 4, 9. The shortest distance is going to be the distance between negative 3, 5 and negative 4, 9. So if I use the distance formula, I'll have my final answer. So I'll do the difference in the x coordinates squared. plus the difference in the y coordinates squared. So this gives me 1 plus 16, so I have root 17 units.
So that's the shortest distance from the point to the line. In this unit you look at the three versions of equations a quadratic function can be given to you in standard vertex and factored form. you learn the usefulnesses of all three versions, and how to transition between the three versions.
Now quadratics are equations where the the highest degree exponent on x is a 2, and the shape of those equations is always this u-type shape which we call a parabola. Now a standard form equation is most useful for determining the y-intercept. So the y-intercept we can tell from the equation because it's the c-value.
So this point right here would be the point 0, and then whatever the c-value is in the standard form equation. Another useful thing about standard form is that the a value tells you the direction of opening. If a is positive, the parabola opens up like the one we have here.
If a is negative, it opens down. Now a quadratic function could also be written in vertex form or factored form, and those have different uses. Now the a value in all three versions of the equation all tell you the same thing.
They all tell you direction of opening. So let's just talk about what the other variables tell you. So in vertex form, well it's obviously most useful for telling you what the coordinates of the vertex are. So if it's in vertex form, y equals a times x minus h squared plus k, the vertex is at h k. And notice since the general format is x minus h, this h value is always going to be the opposite sign of what you see in the equation.
So our vertex is at h k, that's what vertex form is useful for. And factored form is useful for the x-intercepts. So these points right here, you could figure those out.
by if we have it in factored form the form y equals a times x minus or times x minus s we can find the x intercepts by looking at the r and s values this x intercept i can say is at r0 and this one is at s0 so the x intercepts are at r and s let's look at how we can work with the three versions of a quadratic and how to transition between the three question 11 this quadratic is in vertex form Complete the table of information. So the vertex is going to be at, right? Because the general format is, so it must be in order for it to look like, which means h is actually.
So my vertex is at. x coordinate is always opposite sine of what you see there. The axis of symmetry is the vertical line that would divide the parabola in half, and that vertical line, notice it passes right through the vertex. So the equation of the axis of symmetry is x equals whatever the x-coordinate of the vertex is, so x equals h.
So in this case, x equals negative 4. Stretch or compression, we get that by looking at the absolute value of a. So the absolute value of negative 2 is 2. So if the absolute value of a is bigger than 1, it causes a vertical stretch. So vertical stretch, we say by a factor of the absolute value of a. So the absolute value of negative 2. is 2. Always use the absolute value when describing the stretch factor. The direction of opening is affected by the sign.
Since the a value is actually negative, that tells us the quadratic is going to open down. Values that x may take also called the domain of the function, the answer for quadra x is always the same. There's no restrictions on the domain, we just say x can be any real number. Values that y may take, well since the parabola opens down, we know that the vertex is going to be at the very top of the parabola, right?
The parabola is going to look like this, the vertex is at the top, so we know the y values are always going to be at or below the y coordinate of the vertex. So the y values are always going to be less than or equal to the y-coordinate of the vertex, which is 3. If I'm going to graph this function, I'm going to want to make a table of values. And in the table of values, you always want to put the vertex in the center. So I'm going to put negative 4, 3 in the center.
And now I want to pick points on either side of the vertex. So to the left of negative 4, I'm going to choose x values negative 5 and negative 6. And then to the right of negative 4, I'll choose x values negative 3 and negative 2. And now I just have to plug those in to my equation for x. and calculate the y's.
So if I plug negative 6 in for x, I would have y equals negative 2 times negative 6 plus 4 squared plus 3. If I evaluate this, negative 2 squared is 4 times negative 2 is negative 8 plus 3 is negative 5. So this would be negative 5. And since I know parabolas are symmetrical, if 2 to the left of the vertex of the y-coordinate is negative 5, 2 to the right, it's going to be at the same spot. Let me plug negative 5 in for x and see what I would get. Negative 5 plus 4 is negative 1, squared is 1, times negative 2 is negative 2, plus 3 is 1. So I know those are both 1. So using those points, I could graph the quadratic.
And there's a rough sketch of the parabola. Let's go the other way. Here's the graph.
Let's find the vertex form equation. So remember, vertex form looks like this. a times x minus h squared, don't forget the squared, plus k. In order to write the equation, well, I just need to know the vertex.
plug that in for h and k, but I also need to know a, the stretch or compression factor. And we'll solve for that by plugging in the vertex and a point x, y. So I've got the vertex here, it's at 2, negative 6, and it gives me another point here, it labels the y-intercept, that's at 0, negative 4. I've got my x, y, my h, k, sub all that in and solve for a.
I've got y value of negative 4, I don't know a, that's what we're solving for, x is 0. h is 2 squared plus my k value of negative 6. So plus negative 6, I'll just write minus 6. Let's solve this for a. So in the brackets, I'll do first 0 minus 2 is negative 2. Square that, I get 4. Now, it often helps, instead of writing a times 4, usually people write that as 4a. If we write it like that, you'll avoid mistakes.
Let me now add the 6 over. Negative 4 plus 6 is 2. So I have 2 equals 4a. Divide both sides by 4. I get a half equals my a value. So my final answer would have a, h, and k plugged into the vertex form.
So a half x minus 2 squared minus 6. There's the vertex form equation. 13 just asked me to describe transformations in words. So we can describe the left and right shift based on where the vertex is. So my vertex is at negative 5, negative 4. I get that from the h and k.
Remember, h is always the opposite sign of what we see there. The vertex of x squared is at 0, 0. But now the vertex of this function is at negative 5, negative 4. So for that to happen, it must have shifted left 5 units and shifted down 4 units. Now, if I look at the absolute value of a, that will tell me the vertical stretch or compression factor.
The absolute value of a is a third. If the absolute value is between 0 and 1, that's going to vertically compress it, so I would say vertical compression by a factor of the absolute value of a, so by a factor of a third. But the fact that it's negative means it's also going to be opening down, which means it's been vertically reflected, so we would say vertical reflection. Example 14. For this quadratic that's in factored form, we want to state the x-intercepts, the vertex, and the axis of symmetry, and use the information to graph it. Well, to find the x-intercepts, if it's in factored form, that's easy.
We're just looking to see what value of x would make any of the factors be 0. What would make x minus 2 be 0 would be if x was 2. So 2 is an x-intercept. So we have an x-intercept at 2, 0. And what value of x would make this factor be 0? Well, 4 would make it be 0. So there's another x-intercept at 4, 0. If we can make one of the factors be 0, we make the whole product be 0, which makes the y-coordinate become 0. So 2 and 4 are both x values that have a y value of 0, meaning they're x intercepts.
And that's easy to pick out from factored form, right? It's just these numbers here. Remember, opposite sign of what you see there.
Now, where's the vertex? Well, I know parabolas are symmetrical, so the vertex is going to be halfway between those. So a shortcut to find the x coordinate of the vertex would be to just find the average of the x intercepts.
So the x coordinate of the vertex would be equal to the average of 2 and 4. And you average numbers by adding them and dividing by 2. So the x-coordinate of the vertex is 3. But where's the y-coordinate? To find the y-coordinate of the vertex, we have to take that x-coordinate and plug it into the original equation. So I would have negative 4 times 3 minus 2 times 3 minus 4, which is negative 4 times 1 times negative 1, which is 4. So my vertex is at the point 3, 4. And the axis of symmetry is the vertical line that goes through the middle of the parabola.
So the parabola looks, I'll just do a rough sketch now, the parabola looks something like this. The axis of symmetry would be the vertical line, and it's going to go right through the vertex. The equation of that line would be x equals the x coordinate of the vertex.
So x equals 3 is the equation of the axis of symmetry. 15. A parabola has x-intercepts of negative 8 and 2 and passes through the point 0, negative 8. Determine the equation of this parabola in the form here, this factored form. Okay, well the x-intercepts are our r and our s, and we know a point, x, y. To write the equation in factored form, we just have to solve for a. So let's sub in what we know.
y is negative 8. I don't know a, but my x value is 0. So I do 0 minus r, so minus negative 8. times zero minus s and s is two. Now let's solve this for a negative eight equals a zero minus negative eight is eight and zero minus two is negative two. So negative eight equals a times eight times negative two. I'll write that as negative 16 a divide both sides by negative 16 to isolate a and I get a equals a half.
So I could write my final equation y equals a half. and then sub in for r and s, but leave x and y. So x minus negative eight, which we would write as plus eight, and x minus two.
There we have it. Question 16, factor each of the following. Part a, I have a three term quadratic, the coefficient of that x squared term is one, which means I can do product and sum factoring the short way. So I just have to find numbers who have a product of the c value, 24. and a sum of the b value 10. The numbers that multiply to 24 and add to 10 are 6 and 4. Since the coefficient of the x square is 1, I can go right to my factored form equation by just adding 6 and 4 to x in both of those factors.
So x plus 6 and x plus 4 are my factors. Part b is very similar. three term quadratic, coefficient of x squared is one.
So I just need to find numbers that have a product of the c value negative 12, and a sum of the b value, which is one. And the numbers that satisfy that product and sum are four and negative three. So once again, what I do is I can go right to my factored form equation by adding those two numbers to the x's in each of these factors. So x plus four and x plus negative three, which we would just write as x minus three. Part c, three term quadratic, but the leading coefficient, the coefficient of the x squared is not one.
So what we should do is we should check if it can common factor out. And it can in this case, two divides evenly into 22 and 48. So let's common factor of the two by dividing all three terms by two and putting it out front. So x squared plus 11 x plus 24 is my second factor.
And now we're just going to look inside the brackets here at this quadratic, and we're going to try and factor that. The coefficient of the x squared is now 1, so we're just looking for numbers that have a product of the c value, 24, and a sum of the b value, 11. The numbers that work are 8 and 3. So we can factor what's in the brackets to x plus 8 times x plus 3. Just don't lose that 2 that was out front at the beginning. Part d, a three-term quadratic that we want to factor. The coefficient of the x squared is not 1, it's 2, but this time we can't common factor out a 2. 2 doesn't divide evenly into 7 and negative 15, so we can't common factor it out, so we have to factor this the long way.
Some teachers may call it by grouping, some by decomposition, or just the long factoring method, but whatever you call it, we're looking for numbers that don't just have a product of the c value now, a product of a times c, so a product of 2 times negative 15 is negative 30. and a sum of the b value, 7. So the numbers that work in this case are 10 and negative 3. And since the leading coefficient is not 1, we can't go right to our factors. What we need to do is split the middle term into 10x minus 3x. 2x squared plus 10x minus 3x minus 15. So I rewrote the equation, but I rewrote 7x as 10x minus 3x. and specifically those two numbers because those are the numbers that satisfy the product and sum and now we factor it by grouping we look at the first two terms and take out a common factor from the first two so i could take out a 2x and that would leave me with x plus 5. and then i look at the last two terms negative 3x minus 15 and i take a common factor factor from those last two terms. I could take a negative 3 from both terms, and that would leave me with, once again, x plus 5. And now that we have a common binomial, they both have that x plus 5, we can factor out that x plus 5, and be left with 2x minus 3. E.
Once again, a quadratic leading coefficient is not 1, and you can't common factor it out. 3 doesn't divide evenly into 23 and negative 8. So we're going to have to factor it the long way by decomposition and grouping. So I want a product of a times c, 3 times negative 8 is negative 24, and a sum of the b value, which is 23. The numbers that satisfy that product and sum are 24 and negative 1. So I need to split the middle term into 24x minus 1x.
And now I factor by grouping. If I look at the first two terms and take out a common factor, I can take out a 3x. and be left with x plus 8. The last two terms I take out a negative 1 and I'm left with x plus 8. I see that common binomial, that's how I know I'm doing it right.
Take out that common binomial and you're left with 3x minus 1 as your second factor. f is a special product, it's a difference of squares. I have an x squared minus a 16, and 16 is a perfect square number, it's a 4 squared.
So if you have a difference of two perfect square numbers, it factors to x minus 4 times x plus 4. g is another difference of squares. I have 4 x squared minus 25. Well, this one's a little trickier because in order to think of the first term as a perfect square number, we'll think of it as a 2x that's being squared, right? If we square the 2 and the x, we get 4x squared. Minus 25 is 5 squared. And now that I see it's a difference of squares, I know it factors to 2x minus 5 times 2x plus 5. Part h is actually what we call a perfect square trinomial, and that's because it's the same number twice that satisfies the product and sum.
The coefficient of this is 1, so I just need to find numbers with a product of c which is 9 and a sum of b which is 6, and the numbers that work are 3 and 3. So this would factor to x plus 3 times another x plus 3. But you would never write it twice, you would just write it as x plus 3 squared. Let's go backwards now instead of factoring. Let's expand example 17 says expand each of the following into standard form. So I'm going to have to do my double distributive property, sometimes called foiling, multiply the first terms, the outside terms, the inside terms, and the last terms. That's why it's called foiling.
And if I write all four of those products, so 4x times x is 4x squared, and then 4x times seven is 28x, negative one times x is negative x. and negative 1 times 7 is negative 7. Collect my like terms. My only like terms are 28x minus 1x, which is positive 27x.
Expanding part b is probably the most common mistake people make in the grade 10 math course. x minus 5 all squared is not just x squared minus 5 squared. You can't put the squared on both terms. You actually have to foil it out. You have to do x minus 5 times another x minus 5. Rewrite it like that, and then do your double distributive property.
and when you do all four products you'll see what you get x times x is x squared x times negative 5 is negative 5x then negative 5 times x is another negative 5x and then negative 5 times negative 5 is positive 25. if i collect my like terms the middle two terms negative 5x minus 5x is negative 10x and there it is expanded Okay, question 18. I want to get these standard form quadratic equations into vertex form quadratic equations. The process for doing that is you start by putting the first two terms in brackets, and then you common factor out the coefficient of the x squared from the first two terms, but it's just one in this first example, so I don't have to do anything. After that, what we do is we add and subtract a special number inside the brackets.
So I need to add and subtract. a certain number inside the brackets here. Now what number goes here and here, you always do half of this number and then square it and then put that in these two spots. So half of six is three, square it and we get nine. So we add nine and subtract nine.
Really we added zero, right? Nine minus nine is zero, but that's the step that creates a perfect square trinomial, which is going to help it factor very nicely into vertex form. Before I can factor it though, I need to get this negative nine out of the brackets. I don't really want it in there. So you have to get it out by multiplying by whatever's in front of the brackets, which in this case is just one.
So really, we're just moving the negative nine out. So I have x squared plus six x plus nine, negative nine moves out. Now I just have two things left to do. Factor this quadratic. So what numbers multiplied to nine?
and add to six are three and three. So it goes to X plus three times another X plus three, which is X plus three squared. We very intentionally created it so that it would do that. And outside the brackets, negative nine plus 11 is positive two. So my vertex is negative three, two.
And since the A value is positive one, I know the parabola opens up, which means the vertex is at the bottom. So the vertex is a min. So the vertex is a min point.
Let's do the question again, but part B, a little more difficult this time. Same steps, put the first two terms in brackets. Step two, whatever the coefficient of the x squared is, common factor it out from the first two terms that are in brackets. So I'll put the three out front, divide both of the terms in brackets by three. I get x squared plus eight x.
And now what we do is inside the brackets, same thing. We want to create a perfect square trinomial by adding and subtracting Half of this 8 squared, right? You always look at this number, take half of it, and then square it.
So half of 8 is 4, 4 squared is 16. So I'm going to add and subtract 16. We don't want that negative 16 in there, so I take it out of the brackets by multiplying it by whatever's in front. So I have y equals 3x squared plus 8x plus 16. Outside of the brackets, this negative 16 is being multiplied by 3, so it's negative 40. 48 and then I just have to factor the quadratic the numbers that multiply to 16 and add to 8 are 4 and 4 so it factors to x plus 4 times another x plus 4 which we write as x plus 4 squared and then if I do negative 48 minus 17 I get negative 65 so my vertex is negative 4 negative 65 and because the a value is positive I know the parabola opens up which means the vertex set the bottom so once again it's a min point. Almost done with quadratics, let's just solve some equations.
So when solving quadratic equations you always want to get it into factored form, set each factor to zero, and solve each of those equations you make. So the first one's a difference of squares, it's an x squared minus 6 squared, so I know that factors to x minus 6 times x plus 6. And how can this equation be true? Is if either of the factors were 0, it would make the whole product be 0, making the whole equation true. So set each factor to 0. Set x minus 6 equal to 0. Set x plus 6 equal to 0. And solve both of those equations you made. So I get a first answer, x is 6. And my second answer is x equals negative 6. These are both answers that satisfy the original equation.
This one. This one's not set to 0 to start with so we should always start by doing that. If you have a quadratic start by setting it to 0. So x squared plus 4x minus 21 equals 0. Now we want to factor it. So I'd be looking for numbers that have a product of negative 21 and a sum of 4 and since the coefficient of the x squared is 1 I'm going to be able to go right to my factors once I find the numbers that satisfy that product and sum. So the numbers that satisfy the product and sum are 7 and negative 3. So they go in those two spots.
And now to find how the product of these two things could be zero is if either of those two things were equal to zero. So set both the factors equal to zero and solve both of those equations. My first answer is negative seven.
My second answer, three. See, it's set to zero. The coefficient of the x squared isn't one, it's negative one, which I don't really like.
So I'm going to common factor that out. So divide all three terms by negative one, it just changes the sign of all of them. And now in the brackets is a quadratic with a leading coefficient of 1, so I would just need to find the numbers that have a product of negative 6 and a sum of negative 5, and put those numbers added to x.
So those numbers would be negative 6 and 1. And now this product is 0 if any of the factors are 0. So set the factors that involve an x equal to 0, and solve each of those equations, and solve for those two cases. So I get 6 and negative 1 for my two answers. This one, start by setting it equal to 0. I can't common factor with that 5, so I'm going to have to factor this the long way. The numbers that multiply to 5 times negative 4, so I have a product of negative 20, and a sum of the b value negative 19, well that'd be negative 20 and 1. So I'll split the middle term into negative 20x plus 1x.
And now I'll take a common factor from the first two terms. I'd take out a 5x, and I'd have x minus 4. From the last two terms, I can just take out a 1, which leaves me with x minus 4. I have that common binomial of x minus 4. When I take that out, I'm left with 5x plus 1. And now I set each factor to 0. and solve each of those equations. I get 4, and this one's a two-step one to solve.
Subtract the 1 over and then divide the 5. My second answer is negative 1 over 5. Two more to solve. Notice the first four I solved by factoring. These ones actually aren't factorable. There's nothing that multiplies to 5 and adds to 7, so we can't factor it. But that doesn't mean there's no solutions, it just means there's no rational solutions.
There may be some irrational ones, so we need quadratic formula to check. So if I use quadratic formula, x equals negative b, so negative 7, plus or minus the square root of b squared, so 7 squared, minus 4 times a times c, and this is all over 2 times a. Now I recommend to simplify underneath the square root first, because we call that the discriminant, and that's going to reveal to us how many answers we're going to get. So 7 squared is 49. minus 20, that's 29. So I have root 29. Okay, since I got a positive discriminant, that means I'm going to get two answers here.
So I'll split them into my two answers. I've got negative seven plus root 29 over two. And I also have x equals negative seven minus root 29 over two.
Okay, I've got my two exact answers. Let's get to approximate answers by evaluating on our calculator and rounding to two decimal places. The first one would give me negative 0.81, and the second one would give me negative 6.19. If we look at f, I mean sometimes we can't get any answers. You'll see what I mean with this question.
This one, not factorable. There's no numbers that have a product of 14 and a sum of 4. So we can't factor it, so we try quadratic formula. x equals negative b, so negative 4, plus or minus the square root of b squared.
minus 4 times a times c all over 2 times a. The discriminant, if we simplify that, 16 minus 56, we actually get a negative answer. We get negative 40. So we've got the square root of negative 40. And you can't square root a negative.
So this means we're not going to get any real solutions for this. So we would say no real solution. We're almost through the whole course.
I know my y-intercept. I'm also going to want the x-intercepts. So I'm going to get this into factored form.
The x-intercepts have a y-coordinate of 0. So I'm going to set y to 0. And now I'm going to solve this. And then that will give me the x-intercepts. So to solve it, I'll have to factor it. What numbers have a product of 12 and a sum of 8?
Those numbers are 6 and 2. So there's factored form. The product would be 0 if either of the factors were 0. So set each factor to 0 and solve. And I get negative 6 and negative 2. So my x-intercepts are at negative 6 and negative 2. Let me find the vertex now.
Well, I don't have to worry about putting this quadratic into vertex form. because I know the vertex is going to be halfway between the x-intercepts because the parabolas are symmetrical. So I can find the x-coordinate of the vertex just by finding the average of the x-intercepts. So add them and divide by 2. Negative 8 over 2 is negative 4. The vertex is going to have an x-coordinate of negative 4. What's the y-coordinate of the vertex? Just plug negative 4 into the original equation.
So it would be negative 4 squared plus 8 times negative 4 plus 12. 16 minus 32 plus 12. Hey, that's negative 4 as well. So my vertex is at negative 4, negative 4. So that's this point right here. I might as well plot one more point just because I know problems are symmetrical.
If 4 units to the right of the vertex were at a y-coordinate of 12, 4 units to the left, 1, 2, 3, 4, we would also be at a y-coordinate of 12. So now I should be able to draw a fairly accurate sketch of this parabola, making sure I go through all of these key points here and have that good U shape, and there we go. Last quadratic's question is an application question. An object is launched upward at 64 feet per second from a platform 80 feet high.
The equation for the object's height and feet is based on time in seconds as given by this equation. When does the object land on the ground? Anything that time you see anything about the ground, that's the x-intercept. That's when the height is zero.
So set the height to zero and solve. And to solve a quadratic that's set to 0, we factor it. So I'm actually going to common factor out this negative 16 first, because I think it divides evenly into 64 and 80. So let's divide all three terms by negative 16, and I would get x squared minus 4x minus 5. And now this quadratic has a leading coefficient of 1, so I just have to find numbers that multiply to negative 5 and add to negative 4, and those numbers are negative 5 and... positive one.
So there's my factored form version of it. And now this whole product would be zero if any of the factors containing an x was zero. So set each factor with an x to zero and solve, I would get five and negative one.
Now keep in mind x in this equation actually stands for time in seconds. So we can't have negative time. So we can reject that answer. And our only answer here is five seconds.
So it'll hit the ground after five seconds. Part B says, what is the max height of the object? Well, the max, anytime you say anything about max or min, it's talking about the vertex.
And what is the max height? It wants to know the y-coordinate of the vertex. That's what we're looking for here. It's going to be easier for us to find the x-coordinate first and then use that to find the max height.
So I can find the x-coordinate of the vertex by finding the average of the x-intercepts. So the x-coordinate of the vertex, I know it's going to be halfway between the x-intercepts. So I just have to add them.
negative 1 and 5, and divide by 2. That gives me 4 over 2, which is 2. So the x-coordinate of the vertex is 2 seconds. So it's going to be at a max at 2 seconds. But what is the max height?
Well, let's find the y-coordinate of the vertex by plugging 2 into the original equation. If I plug it into the original, negative 16 times 2 squared, plus 64 times 2, plus 80. And if I evaluate this, I would get 144. feet. So my vertex is at the point.
The x coordinate is the time, the y coordinate is the height, so the max height is 144 feet. Part says when is the object 100 feet off the ground? So this one we set the original equation, we set the height to 100. And then we need to solve this equation.
So in order to solve a quadratic we have to set it to zero, so let's move that 100 to the other side. 80 minus 100 is negative 20. And now to solve this, I can't common factor out that entire negative 16, because that doesn't divide evenly into 20. But I could common factor out part of it. So zero equals I can take out, let's say a negative four from all three terms, I'll take our negative four, and that would give me four x squared minus 16 x plus five.
And now Are there any numbers that have a product of 4 times 5, so a product of 20 and a sum of negative 16? I don't think so, so we'd have to use quadratic formula to see how this factor could be 0. So quadratic formula with that factor, I would do x equals negative b, so 16, plus or minus the square root of negative 16 squared, minus 4 times a times c, all over 2 times a. Notice this negative 4 is not part of the quadratic equation at all. All we're doing is trying to figure out how can this be 0, because that will make the whole product be 0, which is what we want.
And then if we simplify the discriminant underneath the square root, I would get 176. And if I split this into my two answers, if I do 16 plus root 176 over 8, I get 3.66. And if I do 16 minus root 176 over 8, I get 0.34. I get two positive answers.
So I get two times when it's at 100 feet. And if we think about it, if we think about what it looks like, we know it has a y-intercept of 80. That's how high the platform is. And then, of course, it's going to go up and come back down.
The highest it gets, we figured out, was 144. This was at 2144. And this is at... 0.80. So it must be at a height of 100 at two other times, at 0.34 and at 3.66, and that's what we found. In this unit you start off by looking at similar triangles, which are triangles that have the same shape but are different sizes. We know similar triangles have equivalent ratios of sides.
That naturally extends to right angle trigonometry and the acronym SOHCAHTOA, which tells us about ratios of sides in similar right angle triangles. If we look at this right angle triangle here, we know that if a right angle triangle has a reference angle of theta, all other right angle triangles with that same angle theta are similar. They're the same shape, but they may be different sizes.
But because they're similar, we know they have equivalent ratios of corresponding sides. Since a triangle has three sides, there's three different pairs we could make from those three sides. We could pair up opposite and hypotenuse to get a ratio, and we call that ratio sine.
That's where so comes from. Sine of an angle equals opposite over hypotenuse. If we pair up the adjacent and hypotenuse, we call that ratio cosine.
Cosine of angle is adjacent over hypotenuse. and tan of an angle is opposite over adjacent and the sides are being labeled from this reference angle theta so of course opposite from that would be that side adjacent means right beside it which is right there and the hypotenuse is always across from the right angle so here's a diagram just illustrating that notice this right angle triangle i have the sine cos and tan ratios being calculated for this 40 degree angle notice if i keep the shape of this the same so keep that 90 degree angle and that 40 degree reference angle but change the size of this triangle notice that these three ratios are all going to stay exactly the same so even though the size of the triangle changes because all of these triangles are similar their ratios stay the same and your calculator has the ratios programmed in so sohcahtoa is used for right angle triangles so let's do a couple questions where we're going to have to solve for angles or sides in right angle triangles so i've got this right angle triangle here i see a reference angle 41 degrees and it's asking me to solve for the indicated side indicated by x so i need the length of side x and i know this side here so from the 41 degree angle i want the side that is opposite to it and i know the hypotenuse so what ratio has opposite and hypotenuse that's sine so i know sine of the 41 degree angle would equal the opposite side which we have called x over the hypotenuse which is 30. and then we'll just isolate x to solve for it by multiplying the 30 over So 30 multiplied by sine of 41 degrees equals x. And if we evaluate that, 30 times sine 41, making sure your calculator's in degree mode, we'll get an approximate answer for the length of side x.
And it's about 19.68. And our units for this are meters. Part B, once again, we have a right angle triangle where we're looking for a side length. So we can use SOHCAHTOA. Here's our reference angle.
from that reference angle we know the opposite side and we are looking for the adjacent side the ratio that has opposite and adjacent is tan so i know tan from the reference angle would equal opposite which is 8 over adjacent which is x now when the variable is in the denominator of the ratio when you do the algebra to isolate it really what happens is the x and the tan 27 just switch spots so i get x equals 8 divided by tan of 27 degrees and i can get my approximate value for that by typing it on my calculator and i get 15.7 meters this one we're looking for an angle when we have a right triangle and we know two sides and we want an angle we actually end up using inverse trig ratios let me show you so from the angle we want we know the opposite side and we know the adjacent side so the ratio that has opposite and adjacent is tan so i know tan of that angle would equal opposite 11 over adjacent 7. to get the angle i actually have to do inverse tan so theta is equal to inverse tan which we denote as tan with that negative 1 looking exponent thing but it's not actually an exponent just means inverse tan of the ratio 11 over 7. and so if we evaluate this inverse tan of 11 over 7 when you use inverse tan The calculator knows your input is the ratio and it's going to output the angle. We do that, we'll get an angle of about 57.53 degrees if we round it to two decimal places. Let's try another one like that.
Once again, a right angle triangle. We know two sides, we're looking for an angle. So from the angle we want, we know the opposite side and the hypotenuse.
The ratio that has opposite and hypotenuse is sine. So I know sine from that angle would equal 51 over 54. and when we want the angle and we know the ratio we do inverse sign of the ratio and the calculator will output to us the angle that has that ratio and it'll be about 70 point eight one degrees let's move on to non right angle trigonometry so after you learned about SOH CAH TOA you would have then learned about two laws that work for calculating sides and angles of non right angle triangles called oblique triangles and you would have learned about sine law and cosine law now sine law and cosine law also would have worked for these first four but SOH CAH TOA is so much quicker that if you have a right triangle you do SOH CAH TOA If you have a non-right triangle, SOHCAHTOA does not work. You have to use sine law and cosine law.
So depending on what you're given and what you're looking for, you decide what to use. So for this question, we know two angles and one side of the triangle, which means we're going to use sine law to find the other side. We're looking for side A, which is of course across from angle A.
Sine law tells us that the ratio of a side to the sine of its opposite angle is equal for each of the sides in its corresponding opposite angle. So I know that A... over sine of its opposite angle, 54, would be equal to 13 divided by sine of its opposite angle, 67. So based on this, I now have an equation with one variable. I can just multiply this sine 54 to the other side, and I then have A isolated. So 13 times sine 54 over sine 67. If I evaluate this, I'll get an approximate length for side A.
which is 11.43 centimeters. Part f, if we're looking for an angle and we know all three sides, we can actually use cosine law for this. So if I want cosine of angle p, it's equal to, we do the square of the side opposite from the angle we want, so 6 squared, minus the squares of the other two sides, so minus 7 squared minus 5 squared. The order of the 7 and the 5 doesn't matter. But the 6 definitely has to be first.
The first one is the one across from the angle we want. Divided by negative 2 times the second two sides, the 7 and the 5. Okay, so we have the ratio. Maybe we want to simplify this ratio.
So 6 squared minus 7 squared minus 5 squared is negative 38. And then negative 2 times 7 times 5 is negative 70. So I have the ratio and I want the angle. I know to get the angle, if I want angle P, I can do inverse cosine of the ratio. And I'll just cancel out those two negatives and just write 38 over 70. So angle P, if I do inverse cos of that ratio, I get 57.12 degrees.
And the last scenario you'll have is if you know two sides and the angle contained by those two sides, which means between the two sides, you can find the side opposite from the contained angle by doing cosine law, a rearranged version of cosine law compared to this one we did define the angle, but it's the same equation. So if I want this side s, the equation based on cosine law is s squared equals the sum of the squares of the other two sides, 7 squared plus 10 squared, minus 2, times those two sides we know, 7 and 10, times cosine of the angle contained by those two sides, 54. So let me simplify the right side of this equation. 49 plus 100 is 149, minus 2 times 7 times 10 is 140, cos 54. Now let me get an approximate value for the right side.
I get 66.71006468. That's what s squared is equal to. I kept all those decimal places because we shouldn't round until the end.
s. would be approximately equal to the square root of this. And if I square root that, round it to two decimal places, it's about 8.17 kilometers.
All right, so that's it for trigonometry. If you have a right angle triangle, use SOHCAHTOA. If you have an oblique triangle, which means a non-right angle triangle, you need to use sine law and cosine law.