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In the figure, CDE is a straight line and A, B, C and D are points on the circle. ∠BCD = 44°, find the value of x.(a) 44° (b) 68° (c) 90° (d) 56° |
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Answer» (c) 90° ∠CDB =\(\frac{1}{2}\) (180º - 44º) = \(\frac{1}{2}\) x 136º = 68º (∵ BCD is an isos. Δ) ∠BAD = 180° – 44° = 136° (opp.∠s of a cyclic quad. are supp.) ∴ ∠ADB = \(\frac{1}{2}\) (180º -136º) = \(\frac{1}{2}\) x 44º = 22º (∵BAD is an isos. Δ) ∴ ∠ADC = ∠ADB + ∠BDC = 22° + 68° = 90° \(\Rightarrow\) x = ∠ADE = 180°–∠ADC =180° – 90° = 90° (∵ EDC is a st. line) |
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