Question

In a right $△\text{ABC}$, right angled at $\text{C}$, if $\angle \text{B}={60}^{\circ }$ and $\text{AB}=15$ units. Find the remaining angles and sides.

Answer 1

Given:

$△ABC$ is right angled at $C,\angle B={60}^{\circ }$

$\angle C={90}^{\circ }$

We know that,

In triangle $ABC$, $\angle A+\angle B+\angle C={180}^{\circ }$

${60}^{\circ }+\angle A+{90}^{\circ }={180}^{\circ }$

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

We have,

$\sin A=\frac{CB}{AB}$

$\sin {30}^{\circ }=\frac{CB}{15}$

$\frac{1}{2}\times 15=CB$

$CB=7.5\text{units}$

From Pythagoras theorem,

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

${15}^2=A{C}^2+{\left(\frac{15}{2}\right)}^2$

$A{C}^2=225-\left(\frac{225}{4}\right)$

$A{C}^2=225\left(\frac{3}{4}\right)$

$AC=\frac{15\sqrt{3}}{2}\text{units}$