Question : In $\triangle$PQR, the angle bisector of $\angle$P intersects QR at M. If PQ = PR, then what is the value of $\angle$PMQ?
Option 1: 75°
Option 2: 80°
Option 3: 70°
Option 4: 90°
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Correct Answer: 90°
Solution : In $\triangle$PQR, PQ = PR ⇒ $\angle$Q = $\angle$R = b PM is the angle bisector of $\angle$P. $\angle$QPM = $\angle$RPM = a Apply angle sum property in $\triangle$PQR, ⇒ $\angle$P + $\angle$Q + $\angle$R = 180° ⇒ $\angle$P + b + b = 180° ⇒ $\angle$P = 180° – 2b $\angle$QPM = $\frac{180° – 2b}{2}$ = 90° – b In $\triangle$PQM, Let $\angle$PMQ = $\theta$ ⇒ $\angle$QPM + $\angle$Q + $\angle$M = 180° ⇒ 90° – b + b + $\theta$ = 180° ⇒ $\theta$ = 90° Hence, the correct answer is 90°.
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Question : In $\triangle \mathrm{PQR}, \angle \mathrm{P}=46^{\circ}$ and $\angle \mathrm{R}=64^{\circ}$.If $\triangle \mathrm{PQR}$ is similar to $\triangle \mathrm{ABC}$ and in correspondence, then what is the value of $\angle \mathrm{B}$?
Question : In a triangle $\triangle \mathrm{PQR}$, the bisectors of $\angle \mathrm{P}$ and $\angle \mathrm{R}$ meet at a point $\mathrm{M}$ inside the triangle. If the measurement of $\angle P M R=127^{\circ}$, then the measurement of $\angle Q$ is:
Question : In a $\triangle ABC$, $\angle A +\angle B=70°$ and $\angle B+\angle C=130°$.Then, the value of $\angle A$ is:
Question : In $\triangle {PQR}, {PN}$ is the median on ${QR}$. If ${PN}={QN}$, then what is the value of $\angle {QPR}$?
Question : O is the incentre of the $\triangle \mathrm{PQR}$. If $\angle \mathrm{POR}=120°$, then what is the $\triangle \mathrm{PQR}$?
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