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In the figure (i) given below, AB is a diameter. The tangent at C meets AB produced at Q, ∠CAB = 34°. Find :

(i) ∠CBA

(ii) ∠CQA

In the figure (i) given below, AB is a diameter. The tangent at C meets AB produced at Q, ∠CAB = 34°. Find (i) ∠CBA (ii) ∠CQA. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Circles

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Answer

From figure,

∠ACB = 90 (∵ angles in semicircle is equal to 90.)

Since sum of angles in a triangle = 180.

In △ABC,

⇒ ∠CAB + ∠ACB + ∠CBA = 180°
⇒ 34° + 90° + ∠CBA = 180°
⇒ 124° + ∠CBA = 180°
⇒ ∠CBA = 180° - 124°
⇒ ∠CBA = 56°.

Hence, the value of ∠CBA = 56°.

(ii) From figure,

∠BCQ = ∠CAB = 34°. (∵ angles in alternate segments are equal.)

∠ACQ = ∠ACB + ∠BCQ = 90° + 34° = 124°.

Since sum of angles in a triangle = 180°.

In △ACQ,

⇒ ∠CAQ + ∠ACQ + ∠CQA = 180°
⇒ 34° + 124° + ∠CQA = 180°
⇒ 158° + ∠CQA = 180°
⇒ ∠CQA = 180° - 158°
⇒ ∠CQA = 22°.

Hence, the value of ∠CQA = 22°.

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