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Question Number 186667 by cherokeesay last updated on 08/Feb/23

Answered by som(math1967) last updated on 08/Feb/23

$$cos30=\frac{{QR}^2+{QT}^2-{RT}^2}{2\times QR\times QT}$$$$\frac{\sqrt{3}}{2}=\frac{16+16-{RT}^2}{2\times 16}$$$$\Rightarrow RT=\sqrt{32-16\sqrt{3}}=4\sqrt{2-\sqrt{3}}$$$$RT=RU=4\sqrt{2-\sqrt{3}}$$$$AreaofA=\frac{1}{2}\times 4\times 4\times sin30=4squnit$$$$\mathrm{\angle }SRU=360-(90+90+75)=105$$$$AreaofB=\frac{1}{2}\times 4\times 4\sqrt{2-\sqrt{3}}\times sin105$$$$=8\sqrt{2-\sqrt{3}}\times \frac{\sqrt{3}+1}{2\sqrt{2}}$$$$=8\times \frac{\sqrt{3}-1}{\sqrt{2}}\times \frac{\sqrt{3}+1}{2\sqrt{2}}=2\times 2=4squnit$$$$\therefore \frac{A}{B}=\frac{4}{4}=1$$

Commented by som(math1967) last updated on 08/Feb/23

Answered by mr W last updated on 08/Feb/23

Commented by mr W last updated on 08/Feb/23

$$foranycase:$$$$A=\frac{ab\sin \theta }{2}$$$$B=\frac{ab\sin (\pi -\theta )}{2}=\frac{ab\sin \theta }{2}=A$$$$\Rightarrow \frac{A}{B}=1✓$$

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