Question

Question 23

In the given figure, PQ = PR, RS = RQ and $ST\parallel QR$. If the exterior $\angle RPU$ is ${140}^{\circ }$, then the measure of $\angle TSR$ is

a) ${55}^{\circ }$
b) ${40}^{\circ }$
c) ${50}^{\circ }$
d) ${45}^{\circ }$

Answer 1

Here,

$\Rightarrow \angle 1={180}^{\circ }-{140}^{\circ }\Rightarrow \angle 1={40}^{\circ }$
Since, PQ = PR
$\therefore \angle Q=\angle R=x$ [say]
In $\Delta PQR,\angle P+\angle Q+\angle R={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow {40}^{\circ }+x+x={180}^{\circ }\Rightarrow 2x={180}^{\circ }-{40}^{\circ }\Rightarrow 2x={140}^{\circ }\Rightarrow x={70}^{\circ }$
So, $\angle Q=\angle R={70}^{\circ }$
Given that, RS = RQ
$\therefore \angle 2=\angle 3={70}^{\circ }$
In $\Delta SQR,\angle 2+\angle 3+\angle 4={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow {70}^{\circ }+{70}^{\circ }+\angle 4={180}^{\circ }\Rightarrow \angle 4={180}^{\circ }-{140}^{\circ }\Rightarrow \angle 4={40}^{\circ }$
Also, $ST\parallel QR$ [given]
Now, $\angle 4=\angle 6={40}^{\circ }$ [alternate interior angles]
$\therefore \angle TSR={40}^{\circ }$