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Question Number 60283 by mr W last updated on 19/May/19

Commented by mr W last updated on 19/May/19

$$\mathrm{\angle }B=15°,\mathrm{\angle }C=30°,BD=DC$$$$findx=?$$

Answered by ajfour last updated on 19/May/19

$$\mathrm{Let}\mathrm{BD}=\mathrm{CD}=1$$$$\frac{\sin \mathrm{x}}{1}=\frac{\sin 15°}{\mathrm{AD}}\mathrm{\&}\frac{\sin (135°-\mathrm{x})}{1}=\frac{\sin 30°}{\mathrm{AD}}$$$$\Rightarrow \sin (135°-\mathrm{x})=\frac{\sin \mathrm{x}}{2\sin 15°}$$$$\Rightarrow 2\sin 15°\left(\frac{\cos \mathrm{x}+\sin \mathrm{x}}{\sqrt{2}}\right)=\sin \mathrm{x}$$$$2\left(\frac{\sqrt{3}-1}{2\sqrt{2}}\right)\left(\frac{\cos \mathrm{x}+\sin \mathrm{x}}{\sqrt{2}}\right)=\sin \mathrm{x}$$$$\Rightarrow \cot \mathrm{x}+1=\frac{2}{\sqrt{3}-1}$$$$\Rightarrow \cot \mathrm{x}=\sqrt{3}\Rightarrow \mathbf{x}=30°.$$

Commented by mr W last updated on 19/May/19

$$thanksalotsir!$$

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