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Question Number 49725 by behi83417@gmail.com last updated on 09/Dec/18

Commented by behi83417@gmail.com last updated on 09/Dec/18

$$ABC,isequilaterl.\frac{AD}{DB}=\frac{1}{2}$$$$\Rightarrow \measuredangle ADC=?$$

Answered by ajfour last updated on 09/Dec/18

$$\frac{AD}{\sin \mathrm{\angle }ACD}=\frac{AC}{\sin \mathrm{\angle }ADC}$$$$\Rightarrow \frac{a/3}{\sin \alpha }=\frac{a}{\sin \theta }\Rightarrow \sin \theta =3\sin \alpha$$$$\frac{BD}{\sin \mathrm{\angle }BCD}=\frac{BC}{\sin \mathrm{\angle }BDC}$$$$\Rightarrow \frac{2a/3}{\sin (\pi /3-\alpha )}=\frac{a}{\sin \theta }$$$$\Rightarrow \sin \theta =\frac{3}{2}\sin (\frac{\pi }{3}-\alpha )$$$$\Rightarrow 2\sin \theta =3(\frac{\sqrt{3}}{2}\cos \alpha -\frac{1}{2}\sin \alpha )$$$$4\sin \theta =3\sqrt{3}\sqrt{1-\frac{\sin {}^2\theta }{9}}-\sin \theta$$$$\Rightarrow 25\sin {}^2\theta =3(9-\sin {}^2\theta )$$$$\sin \theta =\sqrt{\frac{27}{28}}=\frac{3\sqrt{3}}{2\sqrt{7}}$$$$\Rightarrow \theta =\mathrm{\angle }ADC={\sin }^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{7}}\right).$$

Answered by ajfour last updated on 09/Dec/18

$$\sin \mathrm{\angle }ADC=\frac{Altitude\left(CE\right)}{CD}=\frac{\left(\frac{\sqrt{3}a}{2}\right)}{\sqrt{{(\frac{a}{2}-\frac{a}{3})}^2+{\left(\frac{\sqrt{3}a}{2}\right)}^2}}=\frac{3\sqrt{3}}{2\sqrt{7}}$$$$\mathrm{\angle }ADC={\sin }^{-1}\left(\frac{3\sqrt{3}}{2\sqrt{7}}\right).$$

Commented by behi83417@gmail.com last updated on 09/Dec/18

$$nicesir!thanks.$$$$sirAjfour!goodquestionsareuploaded$$$$abovethatyoumayintrestedin.$$

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