Question

Given : sin A = ${\displaystyle \frac{3}{5}}$ , find : (i) tan A (ii) cos A 

Answer 1

Consider the diagram below : 

sin A = ${\displaystyle \frac{3}{5}}$

i .e. ${\displaystyle \frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{3}{5}\Rightarrow \frac{\text{BC}}{\text{AC}}=\frac{3}{5}}$

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 = ${\displaystyle \frac{\text{perpendicular}}{\text{base}}=\frac{3x}{4x}=\frac{3}{4}}$

(ii) cos A = ${\displaystyle \frac{\text{base}}{\text{hypotenuse}}=\frac{4x}{5x}=\frac{4}{5}}$