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: Understanding the Question:
We are given a circle with center $O$.
$PQ$ is tangent to the circle at point $Q$, and $T$ is a point on the tangent line extended past $Q$ such that $P-Q-T$ is a straight line.
The segment $POS$ represents a straight line passing through the center $O$ of the circle, making $QS$ a chord and $OS$ a radius.
We are given $\angle SQT = 55^\circ$ and need to find the measure of $\angle QPS$.

Step 2: Key Formula or Approach:
- The radius of a circle is perpendicular to the tangent at the point of contact: $\angle OQT = 90^\circ$.
- In any triangle, the sum of interior angles is $180^\circ$.
- Radii of the same circle are equal, creating an isosceles triangle $\Delta OQS$ where base angles are equal: $\angle OQS = \angle OSQ$.

Step 3: Detailed Explanation:

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

• Find the angle $\angle OQS$ using the given angle $\angle SQT = 55^\circ$:
\[ \angle OQS = \angle OQT - \angle SQT = 90^\circ - 55^\circ = 35^\circ \]

• In triangle $\Delta OQS$, the sides $OQ$ and $OS$ are equal because they are both radii:
\[ OQ = OS \]
Therefore, the angles opposite to these sides must also be equal:
\[ \angle OSQ = \angle OQS = 35^\circ \]

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

• The angle $\angle PQS$ and the angle $\angle SQT$ lie on the straight tangent line $PT$:
\[ \angle PQS + \angle SQT = 180^\circ \]
\[ \angle PQS = 180^\circ - 55^\circ = 125^\circ \]

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


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