SSC CGL 10 Jun 2019 Shift 1 Paper

Question : 67 of 100

Marks: +1, -0

In ΔPQR, QT ⊥ PR and S is a point on QR such that ∠PSQ = p°. If ∠TQR = 46° and ∠SPR = 32°, then the value of p is :

Solution:

\Δ

PMT

\∠\ {\PMT}=180-(32+90)=58°

\∠\ {\PMT}=\∠ \ {\QMS}

Now, in

\Δ\ {\QMS}

P°=180-(46+58)=76°

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