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Question Number 63663 by Tawa1 last updated on 07/Jul/19

Answered by mr W last updated on 07/Jul/19

$$\mathrm{\angle }ADB=360-(180-6-30)-(180-6-24)=66°$$$$\frac{AD}{DC}=\frac{\sin 30°}{\sin 6°}$$$$\frac{BD}{DC}=\frac{\sin 24°}{\sin 6°}$$$$\Rightarrow \frac{AD}{BD}=\frac{\sin 30°}{\sin 24°}$$$$\frac{AD}{BD}=\frac{\sin (\mathrm{\angle }ADB+x)}{\sin x}=\frac{\sin (66+x)}{\sin x}$$$$\Rightarrow \frac{\sin 30°}{\sin 24°}=\frac{\sin (66+x)}{\sin x}$$$$\Rightarrow \frac{1}{2\sin 24°}=\frac{\sin 66°}{\tan x}+\cos 66°$$$$\Rightarrow \frac{\sin 66°}{\tan x}=\frac{1}{2\sin 24°}-\sin 24°$$$$\Rightarrow \frac{\cos 24°}{\tan x}=\frac{1-2{\sin }^{2}24°}{2\sin 24°}$$$$\Rightarrow \frac{1}{\tan x}=\frac{\cos 48°}{2\sin 24°\cos 24°}$$$$\Rightarrow \frac{1}{\tan x}=\frac{\cos 48°}{\sin 48°}$$$$\Rightarrow \frac{1}{\tan x}=\frac{1}{\tan 48°}$$$$\Rightarrow x=48°$$

Commented by Tawa1 last updated on 07/Jul/19

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

Commented by Tawa1 last updated on 07/Jul/19

$$\mathrm{I}\mathrm{appreciate}\mathrm{sir}$$

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