Question : In a triangle $PQR$, $QR$ is produced to $S$. If $\angle PRS=(9x-15^{\circ}), \angle RPQ=2x$ and $\angle PQR = (4{x}+15^{\circ})$, what is the value of $x$?
Option 1: 55
Option 2: 20
Option 3: 10
Option 4: 75
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Correct Answer: 10
Solution : Given: $\angle PRS=(9x-15^{\circ}), \angle RPQ=2x$ and $\angle PQR=(4{x}+15^{\circ})$ Applying exterior angle property, we get: $\angle PRS=\angle RPQ+\angle PQR$ ⇒ $(9x-15^{\circ}) = 2x+(4{x}+15^{\circ})$ ⇒ $3x=30^{\circ}$ ⇒ $x=10^{\circ}$ Hence, the correct answer is 10.
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Question : The side QR of an equilateral triangle PQR is produced to the point S in such a way that QR = RS and P is joined to S. Then the measure of $\angle PSR$ is:
Option 1: $30^{\circ}$
Option 2: $15^{\circ}$
Option 3: $60^{\circ}$
Option 4: $45^{\circ}$
Question : $\triangle PQR$ and $\triangle SQR$ are both isosceles triangles on a common base $QR$ such that $P$ and $S$ lie on the same side of $QR$. If $\angle QSR=60^{\circ}$ and $\angle QPR=100^{\circ}$, then find $\angle SRP$.
Option 1: $80^{\circ}$
Option 2: $60^{\circ}$
Option 3: $100^{\circ}$
Option 4: $20^{\circ}$
Question : In $\triangle {PQR}, {PN}$ is the median on ${QR}$. If ${PN}={QN}$, then what is the value of $\angle {QPR}$?
Option 1: $90^\circ$
Option 2: $80^\circ$
Option 3: $60^\circ$
Option 4: $75^\circ$
Question : Three angles of a triangle are $(x-15^{\circ}),(x+45^{\circ}),$ and $(x+60^{\circ})$. Identify the type of triangle.
Option 1: Obtuse angle triangle
Option 2: Right angle triangle
Option 3: Isosceles triangle
Option 4: Equilateral triangle
Question : If in $\triangle PQR$ and $\triangle DEF, \angle P=52^{\circ}, \angle Q=74^{\circ}, \angle R=54^{\circ}, \angle D=54^{\circ}, \angle E=74^{\circ}$ and $\angle F=52^{\circ}$, then which of the following is correct?
Option 1: $\triangle \mathrm{PQR} \sim \triangle \mathrm{FED}$
Option 2: $\triangle \mathrm{RQP} \sim \triangle \mathrm{FED}$
Option 3: $\triangle \mathrm{PRQ} \sim \Delta \mathrm{FED}$
Option 4: $\triangle \mathrm{PQR} \sim \triangle \mathrm{DEF}$
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