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Question Number 143909 by mathdanisur last updated on 19/Jun/21

Commented by mr W last updated on 19/Jun/21

$$irememberisolvedthesame$$$$questionsomedaysago.$$

Commented by amin96 last updated on 19/Jun/21

$$?$$

Commented by mathdanisur last updated on 19/Jun/21

$$Sir,ifpossiblepleasetakealookatthis$$

Answered by mr W last updated on 19/Jun/21

Commented by mr W last updated on 19/Jun/21

$$sayBD=DC=1$$$$BC=2\cos 10$$$$\frac{AC}{\sin 40}=\frac{BC}{\sin 60}$$$$AC=\frac{2\cos 10\sin 40}{\sin 60}$$$$\frac{AC}{\sin (x+10)}=\frac{DC}{\sin x}$$$$\frac{\sin (x+10)}{\sin x}=\frac{2\cos 10\sin 40}{\sin 60}$$$$\cos 10+\frac{\sin 10}{\tan x}=\frac{2\cos 10\sin 40}{\sin 60}$$$$1+\frac{\tan 10}{\tan x}=\frac{4\sin 40}{\sqrt{3}}$$$$\tan x=\frac{\tan 10}{\frac{4\sin 40}{\sqrt{3}}-1}$$$$\Rightarrow x={\tan }^{-1}\frac{\tan 10}{\frac{4\sin 40}{\sqrt{3}}-1}=20°$$

Commented by mathdanisur last updated on 19/Jun/21

$$Yes,perfect,thankyoudearsir..$$

Answered by mr W last updated on 19/Jun/21

$$similarquestion:$$

Commented by mr W last updated on 19/Jun/21

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