Question

Please solve this real quick.
I will mark it as a brainlist​

Answer 1

Since BD is a diameter of the circle.

∴∠BAD=90

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and also ∠BCD=90

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∠DBC=∠DCQ=40

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[∠s in the alternate segments]

∠BCP+∠BCD+∠DCQ=180

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⇒∠BCP+90

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+40

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=180

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∠BCP=50

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Similarly from ΔBAD,60

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+∠BAD+∠ADB=180

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⇒60

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+90

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+∠ADB=180

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∠ADB=30

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.

Answer 2

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