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Question Number 139756 by cherokeesay last updated on 01/May/21

Answered by mr W last updated on 01/May/21

Commented by mr W last updated on 01/May/21

$$QB=\sqrt{4^2-2^2}=2\sqrt{3}$$$$\sin \beta =\frac{2}{4}=\frac{1}{2}\Rightarrow \beta =30°$$$$\sin \alpha =\frac{1}{3}\Rightarrow \cos \alpha =\frac{2\sqrt{2}}{3}\Rightarrow \tan \alpha =\frac{1}{2\sqrt{2}}$$$$AQ\sin \beta =(PQ-AQ\cos \beta )\tan \alpha$$$$AQ=\frac{PQ}{\cos \beta +\frac{\sin \beta }{\tan \alpha }}=\frac{6}{\frac{\sqrt{3}}{2}+\frac{2\sqrt{2}}{2}}=\frac{12}{\sqrt{3}+2\sqrt{2}}$$$$AB=2\sqrt{3}-\frac{12}{\sqrt{3}+2\sqrt{2}}=\frac{2(11\sqrt{3}-12\sqrt{2})}{5}\approx 0.833$$

Commented by cherokeesay last updated on 01/May/21

$$thankyoumrW$$

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