Question

Question:-
In the given fig. PQRS is a square and SRT is an equilateral triangle. Find the measure of angle TQR
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Answer 1

Given:-

✯ PQRS is a square.

✯ SRT is an equilateral triangle.

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To Find:-

☆ ∠TQR = ?

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Solution:-

As, PQRS is a square

Therefore, PQ = QR = RS = SP -(i)

∠SPQ = ∠PQR = ∠QRS = ∠RSP = 90°

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As, SRT is an Equilateral triangle

Therefore, RS = RT = TS -(ii)

∠TSR = ∠TRS = ∠STR = 60°

________________________________

From (i) and (ii),

PQ = QR = RS = SP = RT = TS -(iii)

Also, ∠TSP = ∠TSR + ∠SRQ

∠TSP = 60° + 90°

∠TSP = 150° -(iv)

And

∠TRQ = ∠TRS + ∠SRQ

∠TRQ = 60° + 90°

∠TRQ = 150° -(v)

________________________________

From (iv) and (v),

∠TSP = ∠TRQ = 150° -(vi)

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In ∆TSR and ∆TRQ,

TS = TR [from (iii)]

∠TSP = ∠TRQ [from (vi)]

SP = QR [from (iii)]

By SAS congruence criterion,

ΔTSP≅ΔTRQ

Therefore, by CPCT,

PT=QT

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In ΔTQR,

QR=TR [from (iii)]

Therefore, ∠QTR= ∠TQR [angles opposite to equal sides]

________________________________

Now, we know that Sum of angles in a triangle = 180°

Therefore,

∠QTR + ∠TQR + ∠TRQ = 180°

∠TQR + ∠TQR + 150° = 180° [Because we had already proved, ∠QTR = ∠TQR and value of ∠TRQ = 150°]

∠TQR + ∠TQR + 150° = 180°

2∠TQR = 180° - 150°

2∠TQR = 30°

$\sf ∠TQR = \frac{30°}{2}$

$\boxed {\sf ∠TQR = 15°}$

Therefore, value of ∠TQR = 15°.

Answer 2

Answer:

Given:-

✯ PQRS is a square.

✯ SRT is an equilateral triangle.

________________________________

To Find:-

☆ ∠TQR = ?

________________________________

Solution:-

As, PQRS is a square

Therefore, PQ = QR = RS = SP -(i)

∠SPQ = ∠PQR = ∠QRS = ∠RSP = 90°

________________________________

As, SRT is an Equilateral triangle

Therefore, RS = RT = TS -(ii)

∠TSR = ∠TRS = ∠STR = 60°

________________________________

From (i) and (ii),

PQ = QR = RS = SP = RT = TS -(iii)

Also, ∠TSP = ∠TSR + ∠SRQ

∠TSP = 60° + 90°

∠TSP = 150° -(iv)

And

∠TRQ = ∠TRS + ∠SRQ

∠TRQ = 60° + 90°

∠TRQ = 150° -(v)

________________________________

From (iv) and (v),

∠TSP = ∠TRQ = 150° -(vi)

________________________________

In ∆TSR and ∆TRQ,

TS = TR [from (iii)]

∠TSP = ∠TRQ [from (vi)]

SP = QR [from (iii)]

By SAS congruence criterion,

ΔTSP≅ΔTRQ

Therefore, by CPCT,

PT=QT

________________________________

In ΔTQR,

QR=TR [from (iii)]

Therefore, ∠QTR= ∠TQR [angles opposite to equal sides]

________________________________

Now, we know that Sum of angles in a triangle = 180°

Therefore,

∠QTR + ∠TQR + ∠TRQ = 180°

∠TQR + ∠TQR + 150° = 180° [Because we had already proved, ∠QTR = ∠TQR and value of ∠TRQ = 150°]

∠TQR + ∠TQR + 150° = 180°

2∠TQR = 180° - 150°

2∠TQR = 30°

$\sf ∠TQR = \frac{30°}{2}∠TQR= </p><p>2</p><p>30°</p><p> [tex]</p><p> </p><p></p><p>[tex]\boxed {\sf ∠TQR = 15°} </p><p>∠TQR=15°$

Therefore, value of ∠TQR = 15°.