In Fig. 9.4, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to
a. 65°
b. 60°
c. 50°
d. 40°

Solution:

Given, AB is a chord of the circle.

AOC is the diameter of the circle.

Also, ∠ACB = 50°

AT is the tangent of the circle.

We have to find the measure of the angle BAT.

We know that the angle in a semicircle is a right angle.

So, ∠CBA = 90°

We know that the sum of all three interior angles of a triangle is always equal to 180°

In triangle AOB,

∠CAB + ∠CBA + ∠ACB = 180°

∠CAB + 90° + 50° = 180°

∠CAB + 140° = 180°

∠CAB = 180° - 140°

∠CAB = 40° ------------------- (1)

We know that radius of a circle is perpendicular to the tangent at the point of contact.

From the figure,

OA ⟂ OT

So, ∠CAB + ∠BAT = 90°

Substitute (1) in the above expression,

40° + ∠BAT = 90°

∠BAT = 90° - 40°

∠BAT = 50°

Therefore, the measure of the angle BAT is 50°

✦ Try This: In the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10

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NCERT Exemplar Class 10 Maths Exercise 9.1 Problem 3

In Fig. 9.4, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to a. 65°, b. 60°, c. 50°, d. 40°

Summary:

In Fig. 9.4, AB is a chord of the circle and AOC is its diameter such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT is equal to 50°

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