Question

In the given figure, $PQ$ is tangent to the circle with centre $O$. $S$ is a point on the circle such that $\angle SQT = 55^\circ$. The $m\angle QPS$ is

• A. $55^\circ$
• B. $20^\circ$
• C. $35^\circ$
• D. $70^\circ$

Hint:

Using the Alternate Segment Theorem, the angle between the tangent $QT$ and the chord $QS$ ($\angle SQT = 55^\circ$) is equal to the angle subtended by the chord in the alternate segment.
This property often simplifies circle theorems instantly!

Answer 1

The Correct Option is B

Step 1: Set up the right angle at Q.
Since $PT$ is tangent to the circle at $Q$ and $OQ$ is the radius to the point of contact, $OQ \perp PT$, so $\angle OQT = 90^\circ$. Since $\angle SQT = 55^\circ$ and ray $QS$ lies between $QO$ and $QT$, $\angle OQS = 90^\circ - 55^\circ = 35^\circ$.
Step 2: Use the isosceles triangle OQS.
$OQ = OS$ because both are radii, so $\Delta OQS$ is isosceles and its base angles are equal: $\angle OSQ = \angle OQS = 35^\circ$.
Step 3: Apply the exterior angle theorem instead of the angle sum rule.
Since $P, O, S$ are collinear, $\angle QOP$ is the exterior angle of $\Delta OQS$ at $O$, so by the exterior angle theorem it equals the sum of the two remote interior angles: $\angle QOP = \angle OQS + \angle OSQ = 35^\circ + 35^\circ = 70^\circ$.
Step 4: Finish using the right triangle OQP.
In right triangle $OQP$, right-angled at $Q$, the angles sum to $180^\circ$, so $\angle QPO = 90^\circ - 70^\circ = 20^\circ$. Since $S$ lies on line $OP$, this is the same as $\angle QPS$.
\[ \boxed{\angle QPS = 20^\circ} \]