Math, asked by lalithaprasad15, 5 days ago

.If two tangents PA and PB are inclined at an angle of 80^(@) to a circle with centre'O' from a point 'P' find /_POA.​

Answers

Answered by singhsudama579
0

Answer:

50°is the right ans

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Step-by-step explanation:

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

(A) 50° (B) 60° (C) 70° (D) 80°

Solution:

Let's draw a figure as per the 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 (A) 50° (B) 60° (C) 70° (D) 80°

The lengths of tangents drawn from an external point to a circle are equal.

A tangent at any point of a circle is perpendicular to the radius at the point of contact.

In ΔOAP and in ΔOBP

OA = OB (radii of the circle are always equal)

AP = BP (length of the tangents)

OP = OP (common)

Therefore, by SSS congruency ΔOAP ≅ ΔOBP

SSS congruence rule: If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

If two triangles are congruent then their corresponding parts are equal.

Hence,

∠POA = ∠POB

∠OPA = ∠ OPB

Therefore, OP is the angle bisector of ∠APB and ∠AOB

Hence, ∠OPA = ∠ OPB = 1/2 (∠APB )

= 1/2 × 80°

= 40°

By angle sum property of a triangle,

In ΔOAP

∠A + ∠POA + ∠OPA = 180°

OA ⊥ AP (Theorem 10.1 : The tangent at any point of a circle is perpendicular to the radius through the point of contact.)

Therefore, ∠A = 90°

90° + ∠POA + 40° = 180°

130° + ∠POA = 180°

∠POA = 180°-130°

∠POA = 50°

Thus, ans 59°

Answered by ravimdersidhu14
0

Answer:

ans 59

Step-by-step explanation:

I hope this is helpful for you

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