Question
Find $AD$
✓
Answer
In $\triangle ABC ,$
$\angle A C D=\angle A B C+\angle B A C$
$\text { and } \angle A B C=\angle B A C(\because A C=B C)$
$\therefore \angle A B C=\angle B A C=\frac{48^{\circ}}{2}=24^{\circ}$
Now ,
$\frac{A D}{A B}=\sin 24^{\circ}$
$A D=30 \times 0.4067=12.20 m $
$\angle A C D=\angle A B C+\angle B A C$
$\text { and } \angle A B C=\angle B A C(\because A C=B C)$
$\therefore \angle A B C=\angle B A C=\frac{48^{\circ}}{2}=24^{\circ}$
Now ,
$\frac{A D}{A B}=\sin 24^{\circ}$
$A D=30 \times 0.4067=12.20 m $
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