Question

Question 136
PQRS is a rectangle. The perpendicular ST from S on PR divides $\angle S$ in the ratio 2 :3. Find $\angle TPQ$.

Answer 1

Given, $ST\perp PR$ and ST divides $\angle S$ in the ratio 2 : 3.

So, sum of ratio = 2 + 3 = 5
Now, $\angle TSP=\frac{2}{5}\times {90}^{\circ }={36}^{\circ },\angle TSR=\frac{3}{5}\times {90}^{\circ }={54}^{\circ }$

Also, by the angle sum property of a triangle,
$\angle TPS={180}^{\circ }-(\angle STP+\angle TSP)\Rightarrow \angle TPS={180}^{\circ }-({90}^{\circ }+{36}^{\circ })={54}^{\circ }$

We know that, $\angle SPQ={90}^{\circ }$
$\Rightarrow \angle TPS+\angle TPQ={90}^{\circ }\Rightarrow {54}^{\circ }+\angle TPQ={90}^{\circ }\Rightarrow \angle TPQ={90}^{\circ }-{54}^{\circ }={36}^{\circ }$