Question

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In Figure-3, PQ is tangent to the circle with centre at 0, at the point B. If
ZAOB = 100°, then Z ABP is equal to
(A) 50
(C)90
(D)80°
​

Answer 1

Answer:

50° as sum of triangle is 180 and the other two angle are equal so they 40° each and angle at tangent is 90° so 90-40 = 50°

Answer 2

Option (A). 50° is the answer.

Step-by-step explanation:

From the figure attached,

From ΔAOB ≅ 100° ...........[Given]

seg(AO) ≅ seg(OB) ...........[Radius of the circle]

Therefor, m(∠OBA) ≅ m(∠OAB) .........[angles opposite to the equal side of a triangle]

m(∠AOB) + m(∠OAB) + m(∠OBA) ≅ 180°

100° + 2m(∠OBA) = 180°

2m(∠OBA) = 80°

m(∠OBA) = 40°

Since m(∠OBA) + m(∠ABP) = 90°.............[Complementary angles]

∴ m(∠ABP) = 90° - 40°

m(∠ABP) = 50°

Therefore, Option (A) is the answer.

Learn more about theorems of a circle from https://brainly.in/question/1128372