Question
In a right $\triangle ABC$, right angled at $C$, if $\angle B = 60^\circ $ and $AB = 15$ units, find the remaining angles and sides.
✓
Answer
$\angle B=60^{\circ}$
$ \angle C=90^{\circ} \ldots($Since $\triangle A B C$ is right angled at $C)$
$ \angle A+\angle B+\angle C=180^{\circ}$
$ \angle A+60^{\circ}+90^{\circ}=180^{\circ}$
$ \angle A=180^{\circ}-150^{\circ}$
$ \angle A=30^{\circ}$
Now,
$\sin 60^{\circ}=\frac{A C}{A B}$
$ A C=\sin 60^{\circ} \times A B$
$ A C=\frac{\sqrt{3}}{2} \times 15$
$ A C=\frac{15 \sqrt{3}}{2}$ units
Also,
$\cos 60^{\circ}=\frac{ BC }{ AB }$
$ BC =\cos 60^{\circ} \times AB$
$ BC =\frac{1}{2} \times 15$
$ BC =7.5$ units.
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