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

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.