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In figure, $PQ = PR, RS = RQ$ and $ST | | QR$ . If the exterior ∠ $RPU$ is $140^{o}$ , the the measures of ∠ $TSR$ is

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55^{o}

40^{o}

50^{o}

45^{o}

answer is B.

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Detailed Solution

Given,

PQ = PR, RS = RQ

and

ST | | QR

In △

PQR

∠ RPU = ∠ PQR + ∠ PRQ

…(exterior angle property).

∠ PQR = ∠ PRQ

… (angles opposite to equal sides are equal,

PQ = PR

).
⇒

∠ RPU = ∠ PQR + ∠ PQR

⇒

2 ∠ PQR = 140^{o}

⇒

∠ PQR = \frac{140^{o}}{2}

⇒

∠ PQR = 70^{o}

ST is

parallel to

QR

and

QS

is a transversal.
⇒

∠ PQR = ∠ PST = 70^{o}

… (corresponding angles) … (1).

∠ PQR = ∠ SQR = 70^{o}

Now, in △

QSR

∠ SQR = ∠ RSQ = 70^{o}

… (angles opposite to equal sides are equal,

RS = RQ

) … (2).
Here,

PQ

is a straight line.
⇒

∠ PST + ∠ TSR + ∠ RSQ = 180^{o}

… (straight angle).
From equation (

1

) and (

2)

,
⇒

∠ TSR = 180^{o} - 70^{o} - 70^{o}

⇒ ∠

TSR = 40^{o}

Hence, the correct option is 2.

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