Question

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

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

Hint:

An alternative way to solve this is using the Exterior Angle Theorem.
In $\Delta OAP$, the side $PO$ is produced to $B$, so the exterior angle is $\angle AOB = 125^\circ$.
The exterior angle of a triangle is equal to the sum of the two interior opposite angles:
\[ \angle AOB = \angle OAP + \angle APO \]
Substitute the values:
\[ 125^\circ = 90^\circ + \angle APO \implies \angle APO = 125^\circ - 90^\circ = 35^\circ \]
This method requires fewer steps and is less prone to calculation errors.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The topic of this question is Circles and Triangles.
We are given a circle with centre $O$, where $PA$ is a tangent from an external point $P$ to the circle at point $A$.
The line segment $PO$ is extended to meet the circle at point $B$, forming a straight line segment $POB$.
The angle $\angle AOB$ is marked as $125^\circ$ in the diagram (referred to as $\angle POB$ in the question text as an exterior notation).
We need to determine the value of $\angle APO$.

Step 2: Key Formula or Approach:
We will use the following geometric properties:

• The tangent $PA$ is perpendicular to the radius $OA$ at the point of contact $A$. Thus, $\angle OAP = 90^\circ$.

• Angles on a straight line segment add up to $180^\circ$ (Linear Pair Axiom). This allows us to find $\angle AOP$ using the supplementary angle of $\angle AOB$.

• The sum of angles in triangle $OAP$ is $180^\circ$.


Step 3: Detailed Explanation:

• Since $P-O-B$ is a straight line, the angles $\angle AOP$ and $\angle AOB$ form a linear pair on the line segment.

• Therefore, we can write:
\[ \angle AOP + \angle AOB = 180^\circ \]

• Substitute the given value of $\angle AOB = 125^\circ$ into the equation:
\[ \angle AOP + 125^\circ = 180^\circ \]

• Solve for $\angle AOP$:
\[ \angle AOP = 180^\circ - 125^\circ = 55^\circ \]

• Now, look at the right-angled triangle $OAP$. Since $PA$ is a tangent and $OA$ is the radius:
\[ \angle OAP = 90^\circ \]

• The sum of angles in $\Delta OAP$ is:
\[ \angle APO + \angle AOP + \angle OAP = 180^\circ \]

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

• Calculate the final angle:
\[ \angle APO = 180^\circ - 145^\circ = 35^\circ \]


Step 4: Final Answer:
The angle $\angle APO$ is equal to $35^\circ$, which corresponds to option (D).