Question

From the figure, $\sin x^0$ is $\frac{\sqrt{m}}{2}$. The value of $m$ is:

Answer 1

Let the given triangle be ABC, where $\angle ABC={90}^{\circ }$, $AB=1,BC=y,AC=2$
By Pythagoras Theorem,
$A{C}^2=A{B}^2+B{C}^2$
$2^2=1^2+y^2$
$y^2=3$
$y=\sqrt{3}$
$\sin x^{\circ }=\frac{BC}{AC}=\frac{\sqrt{3}}{2}$