Question

If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of $80^\circ$ in the fig, then $\angle POA$ is equal to :

• A. $50^\circ$
• B. $60^\circ$
• C. $70^\circ$
• D. $80^\circ$

Hint:

When two tangents are drawn from an external point to a circle, the line joining the center to that external point bisects both the angle between the tangents and the angle subtended by the chord of contact at the center. Also, the radius is perpendicular to the tangent at the point of contact.

Answer 1

The Correct Option is A

Step 1: Analyze the given information and properties of tangents to a circle.
PA and PB are tangents from an external point P to a circle with center O.
The angle between the tangents, $\angle APB$, is given as $80^\circ$. We need to find $\angle POA$. Key properties of tangents from an external point:
The lengths of tangents from an external point to a circle are equal (PA = PB).
The line segment from the center to the external point (OP) bisects the angle between the tangents ($\angle APB$).
The radius drawn to the point of tangency is perpendicular to the tangent (OA $\perp$ PA and OB $\perp$ PB). This means $\angle OAP = \angle OBP = 90^\circ$.
Step 2: Use the property that OP bisects $\angle APB$.
Since OP bisects $\angle APB$, we have:
$\angle OPA = \frac{1}{2} \angle APB$
$\angle OPA = \frac{1}{2} \times 80^\circ$
$\angle OPA = 40^\circ$
Step 3: Consider triangle $\triangle OAP$.
We know that $\triangle OAP$ is a right-angled triangle because $\angle OAP = 90^\circ$ (radius is perpendicular to the tangent at the point of contact).
The sum of angles in a triangle is $180^\circ$. So, in $\triangle OAP$: $\angle OAP + \angle OPA + \angle POA = 180^\circ$ Substitute the known angle values:
$90^\circ + 40^\circ + \angle POA = 180^\circ$
$130^\circ + \angle POA = 180^\circ$
Step 4: Solve for $\angle POA$.
$\angle POA = 180^\circ - 130^\circ$
$\angle POA = 50^\circ$
Alternatively, consider quadrilateral OAPB:
The sum of angles in a quadrilateral is $360^\circ$.
$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ$
$90^\circ + 80^\circ + 90^\circ + \angle BOA = 360^\circ$
$260^\circ + \angle BOA = 360^\circ$
$\angle BOA = 360^\circ - 260^\circ = 100^\circ$
Since $\triangle OAP \cong \triangle OBP$ (RHS congruence criterion, as OA=OB (radii), OP=OP (common), $\angle OAP=\angle OBP=90^\circ$), OP also bisects $\angle AOB$.
Therefore, $\angle POA = \frac{1}{2} \angle BOA = \frac{1}{2} \times 100^\circ = 50^\circ$.
Step 5: Compare with the given options.
The calculated angle $\angle POA$ is $50^\circ$, which matches option (1). \[ \mathbf{(1)} \quad 50^\circ \]