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Question Number 91760 by  M±th+et+s last updated on 02/May/20

Commented by Tony Lin last updated on 03/May/20

Commented by  M±th+et+s last updated on 03/May/20

$$thanxforsolutions$$

Commented by Tony Lin last updated on 03/May/20

$$\frac{\frac{3}{2}}{x}=\frac{\sqrt{2}}{\sqrt{2+\sqrt{3}}-\sqrt{{\left(\frac{3}{2}\right)}^2+x^2}}$$$$\Rightarrow x=3-\frac{3\sqrt{3}}{2}$$$$\frac{\sqrt{3}}{2}-(3-\frac{3\sqrt{3}}{2})=2\sqrt{3}-3$$$$\therefore \frac{\frac{3}{2}}{3-\frac{3\sqrt{3}}{2}}=\frac{\sqrt{3}}{2\sqrt{3}-3}$$$$\therefore Accordingtoanglebisectortheorem$$$$\Rightarrow 30°splitsintotwo15°=?$$$$?=15°$$

Answered by mr W last updated on 03/May/20

Commented by mr W last updated on 03/May/20

$${AB}^2=1^2+{\left(\sqrt{3}\right)}^2-2\times 1\times \sqrt{3}\cos 120°=4+\sqrt{3}$$$$AB=\sqrt{4+\sqrt{3}}$$$$\cos \mathrm{\angle }CAB=\frac{\sqrt{2}}{\sqrt{4+\sqrt{3}}}$$$$\frac{\sin \mathrm{\angle }DAB}{\sqrt{3}}=\frac{\sin 120°}{\sqrt{4+\sqrt{3}}}$$$$\sin \mathrm{\angle }DAB=\frac{3}{2\sqrt{4+\sqrt{3}}}$$$$x=\mathrm{\angle }CAB-\mathrm{\angle }DAB$$$$={\cos }^{-1}\frac{\sqrt{2}}{\sqrt{4+\sqrt{3}}}-{\sin }^{-1}\frac{3}{2\sqrt{4+\sqrt{3}}}=15°$$

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