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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Question : In an isosceles triangle PQR, $\angle$P = 130°, If I is the in-centre of the triangle, what is the value of $\angle$QIR?
Option 1: 130°
Option 2: 120°
Option 3: 155°
Option 4: 165°
Question : In a $\triangle P Q R, \angle P=90^{\circ}, \angle R=47^{\circ}$ and $P S \perp Q R$. Find the value of $\angle Q P S$.
Option 1: 43°
Option 2: 47°
Option 3: 45°
Option 4: 40°
Question : In the given figure, O is the centre of the circle, $\angle PQR=100^{\circ}$ and $\angle STR=105^{\circ}$. What is the value (in degree) of $\angle OSP$?
Option 1: 95
Option 2: 45
Option 3: 75
Option 4: 65
Question : Two angles of a triangle are equal and the third angle measures 70°. What is the measure of each of the unknown angles?
Option 1: 75°
Option 2: 45°
Option 4: 55°
Question : In the figure, in $\triangle {PQR}, {PT} \perp {QR}$ at ${T}$ and ${PS}$ is the bisector of $\angle {QPR}$. If $\angle {PQR}=78^{\circ}$, and $\angle {TPS}=24^{\circ}$, then the measure of $\angle {PRQ}$ is:
Option 1: 42$^{\circ}$
Option 2: 39$^{\circ}$
Option 3: 30$^{\circ}$
Option 4: 40$^{\circ}$
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