Question

If $\tan A=\sqrt{3}$, find other trigonometric ratios of $\angle A$.

Answer 1

$\tan A=\frac{Perpendicular}{Base}=\frac{\sqrt{3}}{1}$

Using pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$

$A{C}^2=(1{)}^2+(\sqrt{3}{)}^2$

$A{C}^2=1+3=4$

$AC=2$

Therefore,

$Base=1,Perpendicular=\sqrt{3},Hypotenuse=2$

Therefore,

$\sin A=\frac{\sqrt{3}}{2},\cos A=\frac{1}{2},\tan A=\sqrt{3}$

$cosecA=\frac{2}{\sqrt{3}},\sec A=2,\cot A=\frac{1}{\sqrt{3}}$