Question : Circum-centre of $\triangle PQR$ is O. If $\angle QPR=55^{\circ}$ and $\angle QRP=75^{\circ}$, What is the value (in degree) of $\angle OPR$?
Option 1: 45°
Option 2: 40°
Option 3: 65°
Option 4: 70°
Correct Answer: 40°
Solution :
In $\triangle$PQR we have,
$\angle$QPR = 55° and $\angle$QRP = 75°
⇒ $\angle$PQR = 180° – (55° + 75°) = 180° – 130° = 50°
⇒ $\angle$POR = 2 × $\angle$PQR = 2 × 50° = 100°
In $\triangle$OPR,
$\angle$OPR = $\angle$ORP [since OR = OP, both are circumradius]
⇒ $\angle$OPR + $\angle$ORP = 180° – 100° = 80°
⇒ $\angle$OPR + $\angle$OPR = 80°
⇒ 2$\angle$OPR = 80°
⇒ $\angle$OPR = 40°
Hence, the correct answer is 40°.
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