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Question
In the figure, quadrilateral ABCD is cyclic. If m(arc BC) = 90° and ∠DBC = 55°, then find the measure of ∠BCD.
Solution
Given: m(arc BC) = 90°, ∠DBC = 55°
To find: ∠BCD
Solution:
∠BDC = `1/2` m(arc BC) ......[Inscribed angle theorem]
∴ ∠BDC = `1/2 xx 90^circ` ......[Given]
∴ ∠BDC = 45° ......(i)
In ∆BCD,
∠BDC + ∠DBC + ∠BCD = 180° ......[Sum of the measures of all angles of a triangle is 180°]
∴ 45° + 55° + ∠BCD = 180° .....[From (i) and given]
∴ 100° + ∠BCD = 180°
∴ ∠BCD = 180° – 100°
∴ ∠BCD = 80°
∴ The measure of ∠BCD is 80°.
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