Question

ABC is a right triangle, right angled at C. If A = 30${}^{\circ }$ and AB = 40 units, find the remaining two sides and $\angle B$ in $\Delta ABC$.

Answer 1

We have ,
$\angle A+\angle B+\angle C={180}^{\circ }$
$\Rightarrow {30}^{\circ }+\angle B+{90}^{\circ }={180}^{\circ }$ [$\because \angle A={30}^{\circ }and\angle C={90}^{\circ }$]

$\Rightarrow \angle B={180}^{\circ }-{120}^{\circ }={60}^{\circ }$
Now , $cosA=\frac{AC}{AB}$
$\Rightarrow cos{30}^{\circ }=\frac{AC}{40}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{AC}{40}$
$\Rightarrow AC=\frac{\sqrt{3}}{2}\times 40\Rightarrow AC=20\sqrt{3}$ units
and , $sinA=\frac{BC}{AB}$

$\Rightarrow sin{30}^{\circ }=\frac{BC}{40}$
$\Rightarrow \frac{1}{2}=\frac{BC}{40}\Rightarrow =40\times \frac{1}{2}=20$ units

Hence AC = 20$\sqrt{3}$ units , $BC=20$ units and $\angle B={60}^{\circ }$