Question

Use the given figure to find :
(i) sin x^o 
(ii) cos y^o
(iii) 3 tan x^o - 2 sin y^o + 4 cos y^o.

Answer 1

Consider the given figure : 

Since the triangle is a right-angled triangle, so using Pythagorean Theorem 
AD² = 8² + 6²

AD² = 64 + 36 = 100

AD = 10

Also
BC² = AC² – AB²

BC² = 17² – 8² = 225

BC = 15

(i) sin x° = `"perpendicular"/"hypotenuse" = (8)/(17)`

(ii) cos y° = `"base"/"hypotenuse" = (6)/(10) =(3)/(5)`

(iii) sin y° =  `"perpendicular"/"hypotenuse" = "AB"/"AD" = (8)/(10) = (4)/(5)`

    cos y° = `"base"/"hypotenuse" = (6)/(10) = (3)/(5)`

    tan x° = `"perpendicular"/"base" = "AB"/"BC" = (8)/(15)` 

Therefore
3tan x° – 2sin y° + 4 cos y°

= `3 (8/15) – 2 (4/5) + 4 (3/5)`

= `(8)/(5)  – (8)/(5) + (12)/(5)`

= `2(2)/(5)`