Question

If $4{\cos }^{2}A-3=0$ and $0^{\circ }\le A\le {90}^{\circ }$, then find $\angle A$ in degrees.

Answer 1

$4{\cos }^{2}A=3$

${\cos }^2=\frac{3}{4}$

$\cos A=\frac{\sqrt{3}}{2}$ ....(Neglecting the negative value)

$\cos A=\cos 30$

$A={30}^{\circ }$