Question
$PQRS$ is a rectangle. The perpendicular $ST$ from $S$ on $PR$ divides $\angle\text{S}$ in the ratio $2 : 3$. Find $\angle\text{TPQ}$
✓
Answer
Given, $\text{ST}\bot\text{PR}$ and $ST$divides $\angle\text{S}$ in the ratio $2 : 3$
So, sum of ratio $= 2 + 3 = 5$ Now, $\angle\text{TSP}=\frac{2}{5}\times90^\circ=36^\circ,$
$ \angle\text{TSR}=\frac{3}{5}\times90^\circ=54^\circ$
Also, by the angle sum property of a triangle,
$\Rightarrow\text{TSP}=180-(\angle\text{STP}+\angle\text{TSP})$
$\Rightarrow180-(90^\circ+36^\circ)=54^\circ$
We know that, $=180-\angle\text{SPQ}=90^\circ$
$\Rightarrow\angle\text{TPQ}+\angle\text{TPQ}=\ 90^\circ$
$\Rightarrow50^\circ\ \angle\text{TPQ}=90^\circ$
$\Rightarrow\angle\text{TPQ}=90^\circ-54^\circ=36^\circ $
So, sum of ratio $= 2 + 3 = 5$ Now, $\angle\text{TSP}=\frac{2}{5}\times90^\circ=36^\circ,$
$ \angle\text{TSR}=\frac{3}{5}\times90^\circ=54^\circ$
Also, by the angle sum property of a triangle,
$\Rightarrow\text{TSP}=180-(\angle\text{STP}+\angle\text{TSP})$
$\Rightarrow180-(90^\circ+36^\circ)=54^\circ$
We know that, $=180-\angle\text{SPQ}=90^\circ$
$\Rightarrow\angle\text{TPQ}+\angle\text{TPQ}=\ 90^\circ$
$\Rightarrow50^\circ\ \angle\text{TPQ}=90^\circ$
$\Rightarrow\angle\text{TPQ}=90^\circ-54^\circ=36^\circ $
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