Last updated at May 29, 2018 by Teachoo

Transcript

Ex 8.1, 9 In triangle ABC, right-angled at B, if tan A = 1/โ3, find the value of sin A cos C + cos A sin C tan A = 1/โ3 (๐ ๐๐๐ ๐๐๐๐๐ ๐๐ก๐ ๐ก๐ ๐ด)/(๐ ๐๐๐ ๐๐๐๐๐๐๐๐ก ๐ก๐ ๐ด) = 1/โ3 ๐ต๐ถ/๐ด๐ต = 1/โ3 Let BC = x & AB = โ3 x We find AC using Pythagoras theorem Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 AC2 = AB2 + BC2 AC2 = (โ3 ๐ฅ)^2+(๐ฅ)2 AC2 = 3x2 + x2 AC2 = 4x2 AC = โ4๐ฅ2 AC = 2x We need to find sin A , cos A , sin C & cos C sin A = (๐ ๐๐๐ ๐๐๐๐๐ ๐๐ก๐ ๐ก๐ ๐๐๐๐๐ ๐ด)/๐ป๐ฆ๐๐๐ก๐๐๐ข๐ ๐ sin A = ๐ต๐ถ/๐ด๐ถ sin A = ๐ฅ/2๐ฅ sin A = 1/2 cos A = (๐ ๐๐๐ ๐๐๐๐๐๐๐๐ก ๐ก๐ ๐๐๐๐๐ ๐ด )/๐ป๐ฆ๐๐๐ก๐๐๐ข๐ ๐ cos A = ๐ด๐ต/๐ด๐ถ cos A = (โ3 ๐ฅ)/2๐ฅ cos A = โ3/2 We have to find out. sin A cos C + cos A sin C Putting sin A = 1/2 , cos A = โ3/2 , sin C = โ3/2 & cos C = 1/2 = (1/2)ร(1/2)+(โ3/2)ร(โ3/2) = 1/4 + (โ3 ร โ3)/4 = 1/4 + 3/4 = (1 + 3)/4 = 4/4 = 1 So, sin A cos C + cos A sin C = 1 Ex 8.1, 9 In triangle ABC, right-angled at B, if tan A = 1/โ3, find the value of (ii) cos A cos C โ sin A sin C cos A cos C โ sin A sin C Putting sin A = 1/2 , cos A = โ3/2 , sin C = โ3/2 & cos C = 1/2 = (โ3/2)ร1/2โ(1/2)ร(โ3/2) = (โ3/4)โ(โ3/4) = 0 Hence , cos A cos C โ sin A sin C = 0

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.