Advertisements
Advertisements
Question
In the given figure, AB is the diameter. The tangent at C meets AB produced at Q. If ∠CAB = 34°, find:
- ∠CBA
- ∠CQB
Solution
i. AB is a diameter.
∴ ∠ACB = 90°
The angle in a semicircle is the right angle.
∴ In ΔACB,
∠A + ∠C + ∠B = 180°
34° + 90° + ∠B = 180°
∠B = 180° – (34° + 90°)
∠B = 180° – 124°
∠B = 56°
ii. Now,
CQ is tangent.
∴ ∠QCB = ∠CAB ...(Alternate segment angle)
∴ ∠QCB = 34°
And ∠CBQ = 180° – ∠CBA
∠CBQ = 180° – 56° = 124°
∴ ∠CQA = 180° – (∠QCB + ∠CBQ)
∠CQA = 180° – (34° + 124°)
∠CQA = 180° – 158° = 22°.
APPEARS IN
RELATED QUESTIONS
Two circle touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.
From a point P outside a circle, with centre O, tangents PA and PB are drawn. Prove that:
∠AOP = ∠BOP
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:
i)`∠`QPR .
In the following figure, PQ and PR are tangents to the circle, with centre O. If `∠`QPR = 60°, calculate:
- ∠QOR,
- ∠OQR,
- ∠QSR.
In the given figure, AB is the diameter of the circle, with centre O, and AT is the tangent. Calculate the numerical value of x.
AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersects AB produced in D. show that BC = BD.
Tangent at P to the circumcircle of triangle PQR is drawn. If the tangent is parallel to side, QR show that ΔPQR is isosceles.
ABC is a right triangle with angle B = 90°, A circle with BC as diameter meets hypotenuse AC at point D. Prove that: AC × AD = AB2
In Fig. AP is a tangent to the circle at P, ABC is secant and PD is the bisector of ∠BPC. Prove that ∠BPD = `1/2` (∠ABP - ∠APB).
A, B, and C are three points on a circle. The tangent at C meets BN produced at T. Given that ∠ ATC = 36° and ∠ ACT = 48°, calculate the angle subtended by AB at the center of the circle.
