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Ex 8.1, 9 - In ABC, if tan A = 1/ root 3, find sin A cos C - Ex 8.1

Ex 8.1, 9 - Chapter 8 Class 10 Introduction to Trignometry - Part 2
Ex 8.1, 9 - Chapter 8 Class 10 Introduction to Trignometry - Part 3 Ex 8.1, 9 - Chapter 8 Class 10 Introduction to Trignometry - Part 4

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Ex 8.1, 9 - Chapter 8 Class 10 Introduction to Trignometry - Part 5

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Ex 8.1, 9 In triangle ABC, right-angled at B, if tan A = 1/√3, find the value of sin A cos C + cos A sin C tan A = 1/√3 (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝐴)/(𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝐴) = 1/√3 𝐵𝐶/𝐴𝐵 = 1/√3 Let BC = x & AB = √3 x We find AC using Pythagoras theorem Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 AC2 = AB2 + BC2 AC2 = (√3 𝑥)^2+(𝑥)2 AC2 = 3x2 + x2 AC2 = 4x2 AC = √4𝑥2 AC = 2x We need to find sin A , cos A , sin C & cos C sin A = (𝑠𝑖𝑑𝑒 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝐴)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 sin A = 𝐵𝐶/𝐴𝐶 sin A = 𝑥/2𝑥 sin A = 1/2 cos A = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝐴 )/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos A = 𝐴𝐵/𝐴𝐶 cos A = (√3 𝑥)/2𝑥 cos A = √3/2 We have to find out. sin A cos C + cos A sin C Putting sin A = 1/2 , cos A = √3/2 , sin C = √3/2 & cos C = 1/2 = (1/2)×(1/2)+(√3/2)×(√3/2) = 1/4 + (√3 × √3)/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/√3, find the value of (ii) cos A cos C – sin A sin C cos A cos C – sin A sin C Putting sin A = 1/2 , cos A = √3/2 , sin C = √3/2 & cos C = 1/2 = (√3/2)×1/2−(1/2)×(√3/2) = (√3/4)−(√3/4) = 0 Hence , cos A cos C – sin A sin C = 0

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.