Question : Two circles touch externally at P. QR is a common tangent of the circle touching the circles at Q and R. Then, the measure of $\angle QPR$ is:
Option 1: $60^{\circ}$
Option 2: $30^{\circ}$
Option 3: $90^{\circ}$
Option 4: $45^{\circ}$
Correct Answer: $90^{\circ}$
Solution : $\angle$POQ = $\angle$POR = $90^\circ$ OQ = OP = OR (Tangents drawn from the same external point) $\therefore$ $\angle$OQP = $\angle$OPQ = $\angle$ORP = $\angle$OPR In $\triangle$POQ, $\angle$POQ + $\angle$OPQ + $\angle$OQP = $180^\circ$ ⇒ $90^\circ$ + $\angle$OPQ + $\angle$OPQ = $180^\circ$ ⇒ 2$\angle$OPQ = $180^\circ-90^\circ$ $\therefore \angle$OPQ = $45^\circ$ Similarly, $\angle$OPR = $45^\circ$ $\therefore$ $\angle$QPR $=\angle$OPQ + $\angle$OPR $=45^\circ + 45^\circ = 90^\circ$ Hence, the correct answer is $90^\circ$.
Result | Eligibility | Application | Selection Process | Cutoff | Admit Card | Preparation Tips
Question : Two circles touch each other externally at C. AB is a direct common tangent to the two circles, A and B are points of contact, and $\angle CAB = 55^{\circ}$. Then $\angle ACB$ is:
Question : In $\Delta PQR,$ $\angle P : \angle Q : \angle R = 1: 3 : 5$, what is the value of $\angle R - \angle P$?
Question : PT is a tangent at the point R on a circle with centre O. SQ is a diameter, which when produced meets the tangent PT at P. If $\angle$SPT = 32$^\circ$, then what will be the measure of $\angle$QRP?
Question : PQR is a triangle. The bisectors of the internal angle $\angle Q$ and external angle $\angle R$ intersect at S. If $\angle QSR=40^{\circ}$, then $\angle P$ is:
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 $\angle Q P S$.
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile