Mathematics
In the figure (ii) given below, A, B and C are three points on a circle. The tangent at C meets BA produced at T. Given that ∠ATC = 36° and ∠ACT = 48°, calculate the angle subtended by AB at the centre of the circle.
Answer
Join OA, OB and CB. In △ATC,
Ext. ∠CAB = ∠ATC + TCA (∵ external angle in a triangle is equal to the sum of opposite interior angles.)
Ext. ∠CAB = 36° + 48° = 84°.
From figure,
∠ABC = ∠TCA = 48° (∵ angles in alternate segment are equal.)
Since sum of angles in a triangle = 180°.
In △ABC,
⇒ ∠ABC + ∠BAC + ∠ACB = 180°
⇒ 48° + 84° + ∠ACB = 180°
⇒ 132° + ∠ACB = 180°
⇒ ∠ACB = 180° - 132°
⇒ ∠ACB = 48°.
Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.
∴ ∠AOB = 2∠ACB (∵ angle subtended at centre is double the angle subtended at remaining part of the circle.)
∠AOB = 2 × 48° = 96°.
Hence, the angle subtended by AB at the center of the circle is 96°.
Related Questions
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