today we will start chapter 22 trigonometrical ratios for class 9 for ICS from Selena consise for class this is lecture number one of this chapter our trigonometric ratios okay how many ratios are there six six first one is sin Theta second cos Theta third tan Theta fourth C Theta fifth secant Theta and the last one is Con last cos yeah okay sign first and last second and second lastation third and third last sin Theta is equal to equal 1 upon cose Theta cos Theta becomes equal to 1 upon SEC Theta 1 upon secant Theta tan 1 upon C c 1 upon tan secant 1 upon cos and cosecant 1 upon s these are the six trigonometric ratios now 90° angle opposite hypoten hypotenuse remaining two sides base and depute to Theta opposite this side is the perpendicular and third side remains is base change Theta is at here then this side is the perpendicular opposite and third side is base now you have to learn one slogan which is that one Pandit badri pad Sona pad p very good s for sin C4 cos t for Tan so perpendicular upon hypotenuse is s base upon hypotenuse is cos tan P upon B perpendicular upon base oh one upon that is cant cos reciprocal upona 1 upon cos becomes equal to secant and tan reciprocal C base upon perpendicular yes key height hypotenuse hypot and cosecant hypotenuse upon base perpendicular s upon P now sin becomes P upon H cos becomes B upon H tan P upon b c b upon P secant H upon b cose h upon P form first exercise is depend first B second B start exercise exercise 22A first sum is McQ first MC a part if sin a is 5 by3 then what is the value of 10 then a we have to find first of all you have to draw a right angle triangle right angle triangle draw formula perpendicular upon hypotenuse a a accordingly which side is the perpendicular BC BC and the hypotenuse is um AC AC becomes 5 by3 now and let bcal 5K units and AC AC is 13k BC is 5K AC is 13k you can find AB a find [Music] py theor py hypotenuse Square perpendicular square plus b square Now by using py BC squ minus AC squ AC is the hypotenuse or BC is the hypotenuse AC is hypoten AC AC s minus BC s okay hypot square base B hyp perus AC 13 k s BC 5 K 5K squ AB squ 13k square is 169 K squ 5 K square is 25 K squ what we get a is 14 144 k s and a becomes Square become 12 square root 144 square root 12 or k Square very good 12 units formula perpendicular upon base very good which side is the perpendicular here BC BC and Bas is uh base a ab ab BC value 5K or AB ke value 12 K 12 k k k cancel 5 by 12 is our answer first option is the answer second s first car B part 10 a given P upon 5 we same draw a right angle triangle um upon B upon B P Upon a according which side is the BC base 5 AC Square BC sare plus AB very goody theor okay AC Square as like BC BC 3K 3 K Square 3 K Square 9 K Square very good 25 k s AC Square is 34 a² squ and AC becomes square root of 34 k s k is out kot 34 now what they have asked to find they asked to find sin a now sin a formula P upon H which side is a perpendicular BC 3 and hypotenuse AC AC BC AC4 same a base upon hypotenuse base a AB AC can 5 by aunk 34 now we have to find sin square a + cos square a 34 is our LCM 9 + 30 25 sorry 34 by 34 1 is our answer con option is second option is the answer third part you can do it yourself try H same second given base upon perpendicular base value 5K perpendicular value 12K hoty and C A you have to find sare D part in the given figure each obervation is in cim c t c is first of all you have to draw the diagram very good a b c and d this is given AB is given 26 BD is given 10 CD is given 32 32 we have to find tan C tan C we have to find first of all you have to find so a squ becomes equal to AB s minus BD s very good okay 676 minus 100 we get 5766 and Square become square root that is 24 right triang A C tan C tan formula B upon B which side is the perpendicular d 24 and base BC DC not BC DC DC ad value 24 DC Val 348 3x 4 is our answer sear okay e Square Aus tan sare a and t um diam base upon hypotenuse base 3 one answer now we'll reach on question number second uh one more thing B uh 3:34 330 L time come a b C from the following figure find the value of J think you can find AC by py a a t AC B now sin a p upon H second part cos a b upon H third part c a B upon P for secant c h upon B for cose c h upon T the last one is t c p upon P now perpendicular is BC perpendicular is BC four and hypotenuse is uh AB hypoten AC yes base base uh CB CB um AB one five angle depend base AB perpendicular BC CK accordingly which side is the hypotenuse five and base base um four very good perpendicular 3 base 4 these are the answers but is always opposite to right angle next sum third sum figure given this angle is given 90 a b c this side is given 8 this side is given 7 can you find AC yeah how pyus AC Square becomes equal to uh AB squ minus BC squ [Music] uh hypotenuse Square minus perpendicular squar bcus a square 0 289 - 64 15 sides I think so cos B Tan C sin s from the following figure find the value of cos a cose a same right triang know H next sum sixth one sorry fifth one uh same ad find ad or DC yes Lo yes T given a simple s y sin a p upon b h sorry P upon h okay engineering government or private um private okay M housewife mama uh yeah designer ux design okay in the following figure find the value of a oh sorry a sin a first part they have asked to find sin a second part they have asked to find secant a third part cose cos square plus sin square a first of all we have to find AC in term of small a a C Square becomes equal to a square + BC squ AB square + BC Square what is the value of a a a square is a s BC a a square 2 a² what is the value 2 a huh a a root2 sin a p upon H perpendicular a accordingly BC that is a hypoten aun2 1 byun2 but this is the answer now Runk upon 2 second partant H upon B hypotenuse a accordingly aun2 aun and base a aun2 is answer already very good base AB that is a and hypotenuse a [Music] aun2 very now third part they have asked to find COS square a plus sin square a sorry 2 2 2 by 4 is LCM 2 + 2 4X 4 cut and we get one is our answer understood which H eight a given base upon hypotenuse same methodes same a already find same secant a given H upon B hypotenuse 29 K base 21k perpendicular same method given cos a 0.6 find all the trigonometric ratio of angle a COS a 6 by 10 3x 5 in a right right angle triangle it is given that a isle is given sin Theta is given P upon Q draw a right angle triangle name it as ABC Theta is taken at here P upon H perpendicular pyp Theta already given coseta base upon hypotenuse which side is the base this BC upon a k k cancel we get root Q 2 - p² Square upon Q now what they have asked to find they have asked to find COS Theta + sin Theta cos Theta cos thet valus p s upon Q Plus sin Theta sin Theta value P upon q q is the LCM here what we get q^ 2 minus p s plus this is our answer question number 15 is something different cos a half given sin B 1 byun2 given find the value of T A minus tan B upon 1 + t a t b angle A and D are from different right angle triangles triangle draw a b a p q b RM 99 1 by 1 byun2 B upon hypotenuse a 4K minus root3 K same s p upon H perpendicular 1 K hypotenuse < tk2 KK 2k2 2 2 - 1 1 1 k r 1 and bang upon base a accordingly perpendicular root3 K base 1 K cancel root T value tan B same perpendicular upon base 1K upon 1 okay one now what they have asked to find T A minus tan b 1 + t a t b t a value root3 t b value 1 1 upon 1 + < tk3 into 1 < tk3 - 1 1 mp33 < tk3 - 1 whole S A + B A minus B a² - A minus B Square open a sare + b 2 - 2 a 3 - 1 4 2 < tk3 2 say two common 2 - < tk3 2 2 say 2 cancel what we get 2us exercise complete second exercise me 15 something okay yes bye bye