Question

In fig. $\angle ABP=\angle ACQ,\angle BAC={68}^{\circ }$. Measure of $\angle ABP$ is:


[3 Marks]

Answer 1

Given: $\angle ABP=\angle ACQ$
By linear pair property,
$\angle ABP+\angle ABC={180}^{\circ }$ $\cdot \cdot \cdot$ (i)
$\angle ACQ+\angle ACB={180}^{\circ }$ $\cdot \cdot \cdot$ (ii)

From (i) and (ii),
$\angle ABP+\angle ABC=\angle ACQ+\angle ACB$
So, $\angle ABC=\angle ACB$
[$\because$ $\angle ABP=\angle ACQ$ ] [1 Mark]

In $\Delta ABC$,
$\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$
[Using angle sum property of a triangle]
$\Rightarrow 2\angle ABC={180}^{\circ }-{68}^{\circ }$
$[\because \angle ABC=\angle ACB]$
$\Rightarrow 2\angle ABC={112}^{\circ }$
$\Rightarrow \angle ABC={56}^{\circ }$ [1 Mark]

Now, $\angle ABC$ and $\angle ABP$ forms a linear pair, their sum will be ${180}^{\circ }$.
$\Rightarrow$ $\angle ABC$ + $\angle ABP$ = ${180}^{\circ }$
$\Rightarrow \angle ABP={180}^{\circ }-\angle ABC$
$\Rightarrow \angle ABP={180}^{\circ }-{56}^{\circ }$
$\Rightarrow \angle ABP={124}^{\circ }$ [1 Mark]