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 geometric property often simplifies circle theorems instantly!

Answer 1

The Correct Option is B

Step 1: Note that OQ is perpendicular to the whole tangent line.
Since $PQT$ is a single straight tangent line and $OQ$ is a radius drawn to the point of contact, $OQ \perp PQT$ at every point, so $\angle OQP = 90^\circ$ as well as $\angle OQT = 90^\circ$.
Step 2: Find angle QOS using the isosceles triangle OQS.
Since $OQ = OS$ (radii), triangle $OQS$ is isosceles, and $\angle OQS = \angle OQT - \angle SQT = 90^\circ - 55^\circ = 35^\circ$, so $\angle OSQ = 35^\circ$ too. Then $\angle QOS = 180^\circ - 35^\circ - 35^\circ = 110^\circ$.
Step 3: Use the straight line P-O-S to find angle QOP.
Since $P$, $O$, $S$ are collinear, $\angle QOP = 180^\circ - \angle QOS = 180^\circ - 110^\circ = 70^\circ$.
Step 4: Apply the angle sum in triangle OQP.
\[ \angle OQP + \angle QOP + \angle OPQ = 180^\circ \] \[ 90^\circ + 70^\circ + \angle OPQ = 180^\circ \] \[ \angle OPQ = 20^\circ \] Since $O$ lies on $PS$, $\angle QPS = \angle OPQ$.
\[ \boxed{\angle QPS = 20^\circ} \]