Question

Given : sin A = `(3)/(5)` , find : (i) tan A (ii) cos A 

Answer 1

Consider the diagram below : 

sin A = `(3)/(5)`

i .e. `"perpendicular"/"hypotenuse" = (3)/(5) ⇒"BC" /"AC" = (3)/(5)`

Therefore if length of BC = 3x, length of AC = 5x 
Since

AB² + BC² = AC²          ...[ Using Pythagoras Theorem ]
AB² + (3x)² = (5x)²
AB² = 25x² – 9x² = 16x²
∴ AB = 4x (base)

Now
(i) tan A = `"perpendicular"/"base" = (3x)/(4x) = (3)/(4)`

(ii) cos A = `"base"/"hypotenuse" = (4x)/(5x) = (4)/(5)`