Question

Refer to the triangle given in the figure.

$x^0$ in degrees

Answer 1

Given: $\angle ABC={90}^{\circ }$, $AC=20$, $BC=10$ and $AB=y$
By Pythagoras theorem,
$A{C}^2=A{B}^2+B{C}^2$
${20}^2=y^2+{10}^2$
$y^2=400-100$
$y=10\sqrt{3}$
$\cos x^{\circ }=\frac{B}{H}=\frac{BC}{AC}=\frac{10}{20}=\frac{1}{2}$
$\cos x^{\circ }=\frac{1}{2}$
$\cos x^{\circ }=\cos {60}^{\circ }$
$x={60}^{\circ }$