Question 136

Two chords AB, CD of a circle with centre 0 intersect each other at P. ∠ADP = 23° and ∠APC = 70°, then the ∠BCD is

Solution

$$\angle$$APC = $$\angle$$DPB = 70°

=> $$\angle$$APD = 180°-70° = 110° = $$\angle$$BPC

Also, $$\angle$$ADC = $$\angle$$ABC = 23° [Angles in the same segment]

Now, in $$\triangle$$BPC

=> $$\angle$$BCD + $$\angle$$BPC + $$\angle$$PBC = 180°

=> $$\angle$$BCD = 180° - 110° - 23° = 47°

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