Question

In the given figure ,$\Delta ABC$ is equilateral, $\angle P={40}^{\circ }$ and $\angle Q={30}^{\circ }$. Find the value of $x,y$ and $\angle PAQ$, if $PBCQ$ is a straight line.

Answer 1

Consider the given triangle $APB$,

$\angle P={40}^0,\angle Q=30$.

$\angle PBA={180}^{{}^0}-{40}^{{}^0}-x=140-x$ (sum of all angle of triangle$={180}^{{}^0}$ ) ………..(1)

In triangle$ABC,$

$\angle ABC={180}^0-\angle ABP={40}^0-\angle x$ (supplementary angle ) ………..(2)

$\angle BAC=180-\angle P-\angle Q-\angle x+\angle y$

$={180}^0-(40+{30}^0+(\angle x+\angle y))={110}^0-(\angle x+\angle y)$ …………(3)

Given that triangle $ABC$ is equilateral triangle,

Hence, from equation (2) and (3),

$\angle ABC=\angle BAC$

${40}^0-\angle x={110}^0-(\angle x+\angle y)={110}^0-\angle x+\angle y$

${110}^0-\angle y={40}^0$

$\angle y={70}^0$

In triangle ACQ,

$\angle ACQ=180-y-{30}^0={150}^0-\angle y$ …….(sum of all angle of triangle$={180}^{{}^0}$) .......... (4)

Now,

$\angle ACB=180-\angle ACQ$

$\angle ACB={180}^0-{150}^0-\angle y=30-\angle Y$ (supplementary angle) ....... (5)

Given that triangle ABC is equilateral triangle,

Hence, from equation (2) and (5),

$\angle ABC=\angle ACB$

$40-\angle x=30-\angle y$

Put the value of y , we get

$40-\angle x=30-{70}^0$

$40-\angle x=-40$

$\angle x={80}^0$

Hence, $\angle x={80}^0,\angle y={70}^0$

Hence this is the answer.