Mathematics
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
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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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