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

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.

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