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
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