CTET 2 Math and Science Jul 2013 Paper

Question : 36 of 120

Marks: +1, -0

In ΔPQT, PQ = PT. The points R and S are on QT such that PR = PS. If ∠ PTS = 62° and ∠ RPS = 34°, then measure of ∠ QPR is

Solution:

ΔPQT,PQ=PT

∠{PTS}=62^{∘}

and

∠{RPS}=34^{∘}

In ΔPQR and ΔPST
PQ = PT (Given)
PR = PS (Given)

∠PTS=∠PQR

(Angles opposite to equal sides of a triangle are equal) By SAS,

∆{PQR}≅∆{PST}

⇒∠{QPR}=∠{TPS}

.........(i)
Now,In ΔPQT

∠{PTQ}=62^{∘},∠{PQT}=62^{∘}

∴∠{QPT}=180^{∘}-(62^{∘}+62^{∘})=56^{∘}

Again

∠{QPT}=56^{∘}

⇒∠{QPR}+∠{RPS}+∠{TPS}=56^{∘}

[

∵∠{QPR}=∠\mathr{TPS}

from eq(i)]

⇒∠{QPR}+34^{∘}+∠{QPR}=56^{∘}

⇒2∠{QPR}=56^{∘}-34^{∘}⇒2∠{QPR}=22^{∘}

∴∠{QPR}=\;\;{22^{∘}}/{2}=11^{∘}

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