Question

PA is a tangent from an external point P to a circle with centre O. if angle POB =115 then find APO

Answer 1

The angle $\bf \angle APO=25^{\circ}$

Given:

• PA is a tangent from an external point P to a circle with centre O.
• If angle POB =115°.

To find:

• Find $\angle APO$

Solution:

Concept to be used:

• Sum of interior angles of triangle are 180°.
• Tangent make right angle with radius at point of contact.

Step 1:

Draw the figure as attached figure.

As $\angle POB=115^{\circ}$

As AB is a line, so angle formed on this line adds to 180°.

So,

$\angle POB+\angle POA=180^{\circ}$

or

$\angle POA=180^{\circ}-115^{\circ}$

or

$\bf \angle POA=65^{\circ}$

Step 2:

Find Angle APO.

It is clear that $\angle OAP=90^{\circ}$

[Property of tangent]

APO is a triangle.

So,

$\angle AOP+\angle POA+\angle APO=180^{\circ}$

[angle sum property of triangle]

$\angle APO=180^{\circ}-90^{\circ}-65^{\circ}$

or

$\angle APO=25^{\circ}$

Thus,

$\bf \angle APO=25^{\circ}$

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Answer 2

Given:

PA is a tangent from an external point P to a circle with center O.

∠POB = 115°

To Find:

∠APO

Solution:

In ΔAPO,

PA is a tangent from an external point. So, OA and OB are radius of the circle.

So, ∠OAP = 90° [tangent drawn from an external point ⊥ to a circle.]

Since, AB is a line and O is a point on line AB

∠POB +∠POA = 180° [Linear Pair]

Now, ∠POA = 180° -115° [∠POB = 115°(given)]

∠POA =65°

Now In ΔAOP,

by angle sum property a triangle,

∠OAP + ∠ POA + ∠APO = 180°

⇒ 90° + 65° + ∠APO = 180° [∠POA=65° and ∠OAP=90°]

⇒ 155° + ∠APO = 180°

⇒ ∠APO = 180° - 155°

⇒ ∠APO = 25°

Therefore, the measure of ∠APO = 25°.