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Question-224013




Question Number 224013 by efronzo1 last updated on 14/Aug/25
Commented by efronzo1 last updated on 14/Aug/25
Given AB is diameter , ∠AOD=22°   and ∠BCD= p°.   Find p.
$$\mathrm{Given}\:\mathrm{AB}\:\mathrm{is}\:\mathrm{diameter}\:,\:\angle\mathrm{AOD}=\mathrm{22}° \\ $$$$\:\mathrm{and}\:\angle\mathrm{BCD}=\:\mathrm{p}°. \\ $$$$\:\mathrm{Find}\:\mathrm{p}. \\ $$
Answered by A5T last updated on 14/Aug/25
∠ACB=90°; ∠ACD=((22°)/2)=11°  ⇒∠BCD=∠ACB+∠ACD=101°
$$\angle\mathrm{ACB}=\mathrm{90}°;\:\angle\mathrm{ACD}=\frac{\mathrm{22}°}{\mathrm{2}}=\mathrm{11}° \\ $$$$\Rightarrow\angle\mathrm{BCD}=\angle\mathrm{ACB}+\angle\mathrm{ACD}=\mathrm{101}° \\ $$

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