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Question Number 193691 by cortano12 last updated on 18/Jun/23

Commented by cortano12 last updated on 18/Jun/23

Answered by MM42 last updated on 18/Jun/23

$$\mathrm{\angle }PAQ=28\mathrm{\&}\mathrm{\angle }AQP=120$$$$▴APQ:\frac{AP}{sin120}=\frac{10}{sin28}\Rightarrow AP=\frac{5\sqrt{3}}{sin28}$$$$▴ABP:AB=APsin56=\frac{5\sqrt{3}sin56}{sin28}=10\sqrt{3}cos28$$

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