Question

In the figure, $PQR$ is a straight line. Find $x$. Then complete the following :
(i) $\angle AQB=$
(ii) $\angle BQP=$
(iii) $\angle AQR=$

Answer 1

In the given fig., $\angle PQA,\angle AQB$ and $\angle BQR$ forms a linear pair.

Therefore,

$\angle PQA+\angle AQB+\angle BQR=180°$

$(x+20°)+(2x+10°)+(x-10°)=180°$

$4x+20°=180°$

$\Rightarrow x=\frac{180°-20°}{4}=40°$

Therefore,

$\angle AQB=2x+10°=2\times 40°+10°=90°$

$\angle BQP=(x+20°)+(2x+10°)=3x+30°=3\times 40°+30°=150°$

$\angle AQR=(2x+10°)+(x-10°)=3x=3\times 40°=120°$