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: Understanding the Question:
We are given a circle with center $O$.
$PQ$ is a tangent to the circle at point $Q$, and $T$ is a point on the extended tangent line such that $P, Q, T$ lie on a straight line.
The line $PS$ is a secant line that passes through the center of the circle $O$, meaning $QS$ is a chord and $OS$ is a radius lying on the diameter line $POS$.
We are given the angle $\angle SQT = 55^\circ$. We need to find the angle $\angle QPS$.

Step 2: Key Formula or Approach:
- The tangent at any point of a circle is perpendicular to the radius through the point of contact. Therefore, $OQ \perp PT$, which means $\angle OQT = 90^\circ$.
- The triangle $OQS$ is an isosceles triangle because $OQ = OS$ (both are radii of the circle). Hence, the base angles are equal: $\angle OQS = \angle OSQ$.
- In $\Delta PQS$, the sum of the interior angles is $180^\circ$.

Step 3: Detailed Explanation:

• Since $OQ$ is the radius and $PT$ is the tangent line at $Q$:
\[ \angle OQT = 90^\circ \]

• We are given $\angle SQT = 55^\circ$. We can find $\angle OQS$:
\[ \angle OQS = \angle OQT - \angle SQT = 90^\circ - 55^\circ = 35^\circ \]

• Since $OQ$ and $OS$ are both radii of the same circle, we have:
\[ OQ = OS \]

• In $\Delta OQS$, since two sides are equal, the angles opposite to them are also equal:
\[ \angle OSQ = \angle OQS = 35^\circ \]

• Since the points $P, O, S$ lie on a straight line, $\angle OSQ$ is the same as $\angle PSQ$:
\[ \angle PSQ = 35^\circ \]

• The line $PT$ is a straight line, so the angles $\angle PQS$ and $\angle SQT$ are supplementary:
\[ \angle PQS + \angle SQT = 180^\circ \]
\[ \angle PQS + 55^\circ = 180^\circ \]
\[ \angle PQS = 125^\circ \]

• Now, look at the triangle $\Delta PQS$. The sum of angles in $\Delta PQS$ is $180^\circ$:
\[ \angle QPS + \angle PQS + \angle PSQ = 180^\circ \]
\[ \angle QPS + 125^\circ + 35^\circ = 180^\circ \]
\[ \angle QPS + 160^\circ = 180^\circ \]
\[ \angle QPS = 20^\circ \]


Step 4: Final Answer:
The measure of the angle $\angle QPS$ is $20^\circ$.