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In a circle with centre O, a tangent drawn from point X intersects the extended diameter AB at point Y and ∠XYB = 50°, then what is the value of ∠XAB?

In a circle with centre O, a tangent drawn from point X intersects the extended diameter AB at point Y and ∠XYB = 50°, then what is the value of ∠XAB? Correct Answer <span lang="EN-IN" style=" line-height: 107%; background-image: initial; background-position: initial; background-size: initial; background-repeat: initial; background-attachment: initial; background-origin: initial; background-clip: initial;">20°</span>

Given:

∠XYA = 50° and AB is a diameter.

Concept Used:

The angle inscribed in a semicircle is always a right angle (90°).

Sum of all angle in a triangle is 180°.

Alternative segment theorem: In any circle, the angle between chord and a tangent through one of the end point of the chord is equal to the angle in the alternate segment.

Calculation:

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∴ ∠ AXB = 90° (angle inscribed in semicircle)

From alternative segment theorem -

⇒ ∠BAX = ∠BXY = θ

Now, ∠XBA = θ + 50°

In ΔAXB -

⇒ ∠AXB + ∠BAX + ∠XBA = 180°

⇒ 90° + θ + θ + 50° = 180°

⇒ 140° + 2θ = 180°

⇒ 2θ = 40°

⇒ θ = 20°

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