In the figure, $PQRS$ is a square and $SRT$ is an equilateral triangle. Prove that $\angle TQR = 15^o$.
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Given:

$PQRS$ is a square and $SRT$ is an equilateral triangle.

To do:

We have to prove that $PT = QT$.

Solution:

In $\triangle TSP$ and $\triangle TQR$,

$ST = RT$             (Sides of an equilateral triangle)

$SP = PQ$             (Sides of square)

$\angle TSP = \angle TRQ$       

Therefore, by SAS axiom,

$\triangle TSP \cong \triangle TQR$

This implies,

$PT = QT$         (CPCT)

In $\triangle TQR$,

$RT = RQ$                 (Sides of a square)

$\angle RTQ = \angle RQT$

$\angle TRQ = 60^o + 90^o = 150^o$

$\angle RTQ + \angle RQT = 180^o - 150^o = 30^o$

$\angle PTQ = \angle RQT$

This implies,

$\angle RQT = \frac{30^o}{2} = 15^o$

$\angle TQR = 15^o$

Hence proved. 

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Updated on: 10-Oct-2022

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