Question

In the given figure, PA is a tangent from an external point P to a circle with centre O. If $\angle\text{POB} = 125^\circ$, then $\angle\text{APO}$ is equal to :

• A. $25^\circ$
• B. $65^\circ$
• C. $90^\circ$
• D. $35^\circ$

Hint:

Using the exterior angle theorem for $\Delta\text{OAP}$:
The exterior angle $\angle\text{POB}$ is equal to the sum of the two interior opposite angles:
\[ \angle\text{POB} = \angle\text{OAP} + \angle\text{APO} \]
Since $\angle\text{OAP} = 90^\circ$, we have:
\[ 125^\circ = 90^\circ + \angle\text{APO} \implies \angle\text{APO} = 125^\circ - 90^\circ = 35^\circ \]
This method is much faster as it avoids calculating the linear pair!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks us to find the measure of the angle $\angle\text{APO}$ in the given geometric configuration.
We are given a circle with centre O, a tangent line PA from an external point P, and a straight line POB passing through the centre of the circle.
The exterior angle $\angle\text{POB}$ is given as $125^\circ$.

Step 2: Key Formula or Approach:
1. A tangent to a circle is perpendicular to the radius through the point of contact. This gives $\angle\text{OAP} = 90^\circ$.
2. Since POB is a straight line, the angles $\angle\text{AOP}$ and $\angle\text{POB}$ form a linear pair, or we can use the exterior angle property of a triangle.
3. The sum of angles in a triangle is $180^\circ$.

Step 3: Detailed Explanation:

• Identify the properties of the tangent PA at the point of contact A:
The radius OA is perpendicular to the tangent PA.
Therefore, the angle $\angle\text{OAP} = 90^\circ$.

• Since POB is a straight line passing through the centre O:
The angles $\angle\text{AOP}$ and $\angle\text{AOB}$ form a linear pair.
Wait, looking closely at the diagram, the angle $\angle\text{POB}$ is the exterior angle at O for the triangle $\Delta\text{OAP}$.
Therefore, the linear pair relation is:
\[ \angle\text{AOP} + \angle\text{POB} = 180^\circ \]
\[ \angle\text{AOP} + 125^\circ = 180^\circ \]
\[ \angle\text{AOP} = 180^\circ - 125^\circ = 55^\circ \]

• Now, consider the right-angled triangle $\Delta\text{OAP}$:
The sum of all interior angles of $\Delta\text{OAP}$ is $180^\circ$:
\[ \angle\text{OAP} + \angle\text{AOP} + \angle\text{APO} = 180^\circ \]

• Substitute the known values ($\angle\text{OAP} = 90^\circ$ and $\angle\text{AOP} = 55^\circ$):
\[ 90^\circ + 55^\circ + \angle\text{APO} = 180^\circ \]
\[ 145^\circ + \angle\text{APO} = 180^\circ \]

• Solve for $\angle\text{APO}$:
\[ \angle\text{APO} = 180^\circ - 145^\circ = 35^\circ \]


Step 4: Final Answer:
The angle $\angle\text{APO}$ is equal to $35^\circ$.
Hence, option (D) is correct.