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Question Number 169947 by cortano1 last updated on 13/May/22

Commented by cortano1 last updated on 13/May/22

Answered by Rasheed.Sindhi last updated on 13/May/22

$$BD=1\left(say\right)\Rightarrow AC=2$$$$\sin 20=\frac{BC}{AC}=\frac{BC}{2}$$$$BC=2\sin 20$$$$\mathrm{\angle }ACB=90-20=70$$$$\mathrm{\angle }BCD=180-70=110$$$$△BCD:$$$$BD=1,BC=2\sin 20$$$$\frac{BD}{\sin 110}=\frac{CD}{\sin x}=\frac{BC}{\sin (180-110-x)}$$$$\frac{1}{\sin 110}=\frac{CD}{\sin x}=\frac{2\sin 20}{\sin (70-x)}$$$$CD=\frac{\sin x}{\sin 110}$$$$CD=\frac{2\sin 20\sin x}{\sin (70-x)}$$$$\frac{\sin x}{\sin 110}=\frac{2\sin 20\sin x}{\sin (70-x)}$$$$\frac{1}{\sin 110}=\frac{2\sin 20}{\sin (70-x)}$$$$\sin (70-x)=2\sin 20\sin 110$$$$70-x={\sin }^{-1}\left(2\sin 20\sin 110\right)$$$$x=70-{\sin }^{-1}\left(2\sin 20\sin 110\right)$$$$x=30$$

Commented by cortano1 last updated on 13/May/22

$$\frac{1}{\sin 110°}=\frac{2\sin 20°}{\sin (70°-x)}$$$$\Rightarrow \sin (70°-x)=2\sin 20°\cos 20°$$$$\Rightarrow \sin (70°-x)=\sin 40°$$$$\Rightarrow x=30°$$

Answered by mr W last updated on 13/May/22

$$BD=1,AC=2$$$$\tan 20=\frac{\cos x}{2\cos 20+\sin x}$$$$\frac{\sin 20}{\cos 20}=\frac{\cos x}{2\cos 20+\sin x}$$$$\cos x\cos 20-\sin x\sin 20=2\cos 20\sin 20$$$$\cos (x+20)=\sin 40=\cos 50$$$$\Rightarrow x+20=50$$$$\Rightarrow x=30✓$$

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