Question

$PQ=QR=PS$. Calculate the size of the labelled angles.

• A. (a)$a={40}^{\circ },b={50}^{\circ },c={70}^{\circ },d={110}^{\circ }$
• B. (b)$a={42}^{\circ },b={48}^{\circ },c={69}^{\circ },d={111}^{\circ }$
• C. (c)$a={45}^{\circ },b={45}^{\circ },c={67.5}^{\circ },d={112.5}^{\circ }$
• D. (d)$a={50}^{\circ },b={40}^{\circ },c={65}^{\circ },d={115}^{\circ }$

Answer 1

Answer: (C)

(c)$a={45}^{\circ },b={45}^{\circ },c={67.5}^{\circ },d={112.5}^{\circ }$

$\angle Q=90°$ and $PQ=QR$

So, $\angle R=\angle P$

from $△PQR,\angle P+\angle Q+\angle R=180°$

$⟹\angle P=\angle R=45°$

Given $PQ=PS$.

So, $\angle PQS=\angle PSQ,$

$45°+2\angle PQS=180°$

$⟹\angle PQS=\angle PSQ=67.5°$

$d=180°-c°=112.5°$