# If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in Fig. 9.17. Prove that ∠BAT = ∠ACB

**Solution:**

Given, AB is a __chord of a circle__ with centre O.

AOC is a __diameter of the circle__

AT is the tangent at point A

We have to prove that ∠BAT = ∠ACB

We know that angle in a __semicircle__ is always equal to 90°

So, ∠CBA = 90°

Considering __triangle__ CBA,

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

∠CBA + ∠BAC + ∠ACB = 180°

90° + ∠BAC + ∠ACB = 180°

∠BAC + ∠ACB = 180° - 90°

∠BAC + ∠ACB = 90°

∠ACB = 90° - ∠BAC ----------------- (1)

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

So, OA ⟂ AT

∠OAT + ∠CAT = 90°

From the figure,

∠CAT = ∠BAT + ∠BAC

90° = ∠BAT + ∠BAC

∠BAT = 90° - ∠CAB ---------------- (2)

Comparing (1) and (2),

Since RHS are same

∠ACB = ∠BAT

Therefore, it is proved that ∠ACB = ∠BAT

**✦ Try This: **In Fig. PA and PB are tangents to the circle drawn from an external point P. CD is a third tangent touching the circle at Q. If PB = 10 cm and CQ = 2 cm, what is the length of PC?

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

**NCERT Exemplar Class 10 Maths Exercise 9.4 Problem 4**

## If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in Fig. 9.17. Prove that ∠BAT = ∠ACB

**Summary:**

If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in Fig. 9.17. It is proven that ∠BAT = ∠ACB

**☛ Related Questions:**

- Two circles with centres O and O' of radii 3 cm and 4 cm, respectively intersect at two points P and . . . .
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- In Fig. 9.18, tangents PQ and PR are drawn to a circle such that ∠RPQ = 30°. A chord RS is drawn par . . . .

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