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Question Number 47354 by behi83417@gmail.com last updated on 08/Nov/18

Commented by behi83417@gmail.com last updated on 08/Nov/18

$$ABC,isaequelateraltrianglewith$$$$AB=2,and:AC{}^{\prime }=3,{BA}^{\prime }=2.$$$$......\mathbf{CD}=?$$

Commented by ajfour last updated on 09/Nov/18

Commented by ajfour last updated on 09/Nov/18

$$\gamma =30°\Rightarrow \mathrm{\angle }C=90°$$$$A^{\prime }C=\sqrt{4^2-2^2}=2\sqrt{3}...\left(i\right)$$$$From△BCD$$$$\frac{\sin \beta }{2}=\frac{\sin \gamma }{BD}\mathrm{\&}From△{BDA}^{\prime }$$$$\frac{\sin \gamma }{BD}=\frac{\sin (\beta -\gamma )}{A^{\prime }D}$$$$\Rightarrow A^{\prime }D=\frac{3\sin (\beta -\gamma )}{\sin \beta }....\left(ii\right)$$$$From△CC{}^{\prime }D$$$$\frac{\sin \alpha }{CD}=\frac{\cos \beta }{CD}=\frac{\sin \beta }{5}$$$$\Rightarrow \tan \beta =\frac{5}{CD}...\left(iii\right)$$$$Using\left(i\right),\left(ii\right)for$$$$CD+A^{\prime }D=A^{\prime }C$$$$\Rightarrow CD+\frac{3\sin (\beta -\gamma )}{\sin \beta }=2\sqrt{3}...\left(I\right)$$$$letCD=x$$$$Thenusing\left(iii\right)\mathrm{\&}\left(I\right)$$$$x+3(\cos \gamma -\frac{\sin \gamma }{\tan \beta })=2\sqrt{3}$$$$\Rightarrow x+3(\frac{\sqrt{3}}{2}-\frac{x}{10})=2\sqrt{3}$$$$or\frac{7x}{10}=\frac{\sqrt{3}}{2}$$$$\Rightarrow x=CD=\frac{5\sqrt{3}}{7}.$$

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

$$sirAjfour!thankyouverymuch$$$$fornicework.godblessyousir.$$

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