Question
✓
Answer
In the figure,
$A B=A C$ and $B P$ bisects $\angle A B C$
$A P \| B C$ is drawn.
Now $\angle \mathrm{PBC}=\angle \mathrm{PBA}$ ( $\because P B$ is the bisector of $\angle A B C$ )
$\because A P \| B C$
$\therefore \angle \mathrm{APB}=\angle \mathrm{PBC}$ ....... (Alternate angles)
$ \Rightarrow x=\angle \mathrm{PBC} $
In $\triangle \mathrm{ABC}, \angle \mathrm{A}=60^{\circ}$
and $\angle B=\angle C$. $\ldots(\because A B=A C)$
But $\angle A+\angle B+\angle C=180^{\circ}$......... (Angles of a triangle)
$\Rightarrow 60^{\circ}+\angle B+\angle C=180^{\circ}$
$\Rightarrow 60^{\circ}+\angle B+\angle B=180^{\circ}$
$\Rightarrow 2 \angle \mathrm{B}=180^{\circ}-60^{\circ}=120^{\circ}$
$\therefore \angle B=\frac{120^{\circ}}{2}=60^{\circ}$
$\Rightarrow \frac{1}{2} \angle \mathrm{B}=\frac{60^{\circ}}{2}=30^{\circ}$
$\Rightarrow \angle \mathrm{PBC}=30^{\circ}$
$\therefore$ From (i),
$ x=30^{\circ} $
$A B=A C$ and $B P$ bisects $\angle A B C$
$A P \| B C$ is drawn.
Now $\angle \mathrm{PBC}=\angle \mathrm{PBA}$ ( $\because P B$ is the bisector of $\angle A B C$ )
$\because A P \| B C$
$\therefore \angle \mathrm{APB}=\angle \mathrm{PBC}$ ....... (Alternate angles)
$ \Rightarrow x=\angle \mathrm{PBC} $
In $\triangle \mathrm{ABC}, \angle \mathrm{A}=60^{\circ}$
and $\angle B=\angle C$. $\ldots(\because A B=A C)$
But $\angle A+\angle B+\angle C=180^{\circ}$......... (Angles of a triangle)
$\Rightarrow 60^{\circ}+\angle B+\angle C=180^{\circ}$
$\Rightarrow 60^{\circ}+\angle B+\angle B=180^{\circ}$
$\Rightarrow 2 \angle \mathrm{B}=180^{\circ}-60^{\circ}=120^{\circ}$
$\therefore \angle B=\frac{120^{\circ}}{2}=60^{\circ}$
$\Rightarrow \frac{1}{2} \angle \mathrm{B}=\frac{60^{\circ}}{2}=30^{\circ}$
$\Rightarrow \angle \mathrm{PBC}=30^{\circ}$
$\therefore$ From (i),
$ x=30^{\circ} $
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