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° , then ∠POA is equal to.

Answer 1

Step-by-step explanation:

Diagram:- Refer the attachment given above.

Given:-

PA and PB are tangents from a point P.

O is the centre of the circle.

 ∠APB = 80°

To Find:-

The value of  ∠POA

Solution:-

In △AOP and △BOP

⇢  ∠OAP = ∠OBP  【 Radius⊥Tangent 】

⇢ OA = OB 【 Radii 】

⇢ OP = OP 【 Common side 】

∴ △AOP ≅ △BOP 【 By SSA criterion 】

⇢ ∠APO = ∠BPO 【 CPCTC 】

Given, ∠APB = 80°

⇢ ∠APO + ∠BPO = 80°

⇢ ∠APO + ∠APO = 80°

⇢ 2 ∠APO = 80°

⇢ ∠APO = $\dfrac{80}{2}$

⇢ ∠APO = 40°

Since, the sum of all the angles in a triangle is 180°

In △POA ,

⇢ ∠OAP + ∠APO + ∠POA = 180°

⇢ 90° + 40° + ∠POA = 180°

⇢ 130° + ∠POA = 180°

⇢ ∠POA = 180° - 130°

⇢ ∠POA = 50°

$\underline{\boxed{\purple{\rm \therefore \angle POA = 50\degree}}}$

Answer 2

$\huge\mathcal{\fcolorbox{cyan}{black}{\pink{❥ᴀ᭄ɴsᴡᴇʀ࿐}}}$

First, draw the diagram according to the given statement.

Ncert solutions class 10 chapter 10-5

Now, in the above diagram, OA is the radius to tangent PA and OB is the radius to tangents PB.

So, OA is perpendicular to PA and OB is perpendicular to PB i.e. OA ⊥ PA and OB ⊥ PB

So, ∠OBP = ∠OAP = 90°

Now, in the quadrilateral AOBP,

The sum of all the interior angles will be 360°

So, ∠AOB+∠OAP+∠OBP+∠APB = 360°

Putting their values, we get,

∠AOB + 260° = 360°

∠AOB = 100°

Now, consider the triangles △OPB and △OPA. Here,

AP = BP (Since the tangents from a point are always equal)

OA = OB (Which are the radii of the circle)

OP = OP (It is the common side)

Now, we can say that triangles OPB and OPA are similar using SSS congruency.

∴△OPB ≅ △OPA

So, ∠POB = ∠POA

∠AOB = ∠POA+∠POB

2 (∠POA) = ∠AOB

By putting the respective values, we get,

=>∠POA = 100°/2 = 50°

As angle ∠POA is 50° option A is the correct option.