Question

AB is a chord of the circle and AOC is its diameter such that angle ACB = 50°. If AT is the

tangent to the circle at the point A, then BAT is equal to the circle at the point A,then BAT

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Answer 1

Answer:

Step-by-step explanation:

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Answer 2

Given:  ∠ACB = 50°, AC is a diameter passing through the centre of the circle O.

To find:  ∠BAT

Solution:

∠AOC = 180° (∵ AC is a diameter)  

∠ABC =  ½ × ∠AOC  (∵ Measure of angles subtended to any point on the circumference of the circle from the same arc is equal to half of the angle subtended at the centre by the same arc)

∠ABC =  ½ × 180°

∠ABC = 90°

Using Angle Sum Property of a Triangle,

∠ABC + ∠ACB + ∠BAC = 180°

50° +  90° + ∠BAC = 180°  (∵ ∠ACB = 50°,  ∠ABC = 90°)

140° + ∠BAC = 180°

∠BAC = 180° - 140°

∠BAC = 40°

∠CAT = 90° (∵ Tangent at any point of the circle is perpendicular to the radius of the circle through the point of contact)

∠BAT + ∠BAC = ∠CAT (∵Linear Pair)

∠BAT + ∠BAC = 90° (∵ ∠CAT = 90°)

∠BAT + 40° = 90° (∵ ∠BAC = 40°)

∠BAT = 90° - 40°

∠BAT = 50°

∴ The value of ∠BAT is 50°