Question Number 186402 by Rupesh123 last updated on 04/Feb/23
Answered by mr W last updated on 04/Feb/23
Commented by mr W last updated on 04/Feb/23
$$\mathrm{sin}\:\alpha=\frac{\mathrm{4}\:\mathrm{sin}\:\mathrm{75}°}{\mathrm{6}} \\ $$$$\Rightarrow\frac{\mathrm{sin}\:\mathrm{75}°}{\mathrm{sin}\:\alpha}=\frac{\mathrm{6}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\theta=\mathrm{105}°−\alpha \\ $$$$\gamma=\theta−\mathrm{30}°=\mathrm{75}°−\alpha \\ $$$$\frac{{x}}{\mathrm{sin}\:\alpha}=\frac{\mathrm{4}+\mathrm{6}}{\mathrm{sin}\:\theta} \\ $$$$\Rightarrow{x}=\frac{\mathrm{10}\:\mathrm{sin}\:\alpha}{\mathrm{sin}\:\theta} \\ $$$$\frac{{y}}{\mathrm{sin}\:\mathrm{30}°}=\frac{{x}}{\mathrm{sin}\:\gamma} \\ $$$$\Rightarrow{y}=\frac{{x}}{\mathrm{2}\:\mathrm{sin}\:\gamma} \\ $$$${green}\:{area}=\frac{{xy}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$=\frac{\mathrm{25}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{cos}\:\left(\alpha+\mathrm{15}°\right)\:\mathrm{cos}\:\left(\alpha−\mathrm{15}°\right)} \\ $$$$=\frac{\mathrm{25}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{15}−\mathrm{sin}^{\mathrm{2}} \:\alpha} \\ $$$$=\frac{\mathrm{25}}{\left(\frac{\mathrm{sin}\:\mathrm{75}°}{\mathrm{sin}\:\alpha}\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\frac{\mathrm{25}}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{1}}=\mathrm{20}\:\checkmark \\ $$