Question

$sin{x}^{\circ }$ is $\frac{\sqrt{3}}{m}$
value of m is

Answer 1

Given:$\Delta ABC,\angle B={90}^{\circ }$, AB = y units, BC = $\sqrt{3}$ units, AC = 2 units and angle A = $x^{\circ }$,
Using Pythagoras theorem,
$A{C}^2=A{B}^2+B{C}^2$
$2^2=y^2+(\sqrt{3}{)}^2$
$y=1$
Now, $\sin x^{\circ }=\frac{P}{H}=\frac{BC}{AC}=\frac{\sqrt{3}}{2}$