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Question Number 201989 by cortano12 last updated on 18/Dec/23

Answered by mr W last updated on 18/Dec/23

Commented by mr W last updated on 18/Dec/23

$$\sin \alpha =\frac{5}{15}=\frac{1}{3}$$$$\sin \beta =\frac{10}{20}=\frac{1}{2}\Rightarrow \beta =30°$$$${BS}^2=10\times 30\Rightarrow BS=10\sqrt{3}$$$${PB}^2={30}^2+{\left(10\sqrt{3}\right)}^2-2\times 30\times 10\sqrt{3}\times \frac{\sqrt{3}}{2}=300$$$$\Rightarrow PB=10\sqrt{3}=BS\Rightarrow \alpha +\gamma =\beta =30°$$$$\frac{AB}{\sin \gamma }=\frac{PB}{\sin (\alpha +\beta )}$$$$\Rightarrow AB=\frac{10\sqrt{3}\sin (30°-\alpha )}{\sin (30°+\alpha )}$$$$=\frac{10\sqrt{3}(\cos \alpha -\sqrt{3}\sin \alpha )}{\cos \alpha +\sqrt{3}\sin \alpha }$$$$=\frac{10\sqrt{3}(\frac{2\sqrt{2}}{3}-\frac{\sqrt{3}}{3})}{\frac{2\sqrt{2}}{3}+\frac{\sqrt{3}}{3}}=2(11\sqrt{3}-12\sqrt{2})\approx 4.164$$

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