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Question Number 58313 by mr W last updated on 21/Apr/19

Commented by tanmay last updated on 21/Apr/19

Commented by tanmay last updated on 21/Apr/19

$$waitsirtryingtofindx...$$

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

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

$$Method1usingtrigonometry:$$$$\frac{AC}{AD}=\frac{\sin (x+20)}{\sin x}$$$$\frac{AC}{AB}=\frac{\sin 80}{\sin 40}$$$$sinceAB=AD,$$$$\Rightarrow \frac{\sin (x+20)}{\sin x}=\frac{\sin 80}{\sin 40}=2\cos 40=2\sin 50=2\sin (30+20)=\cos 20+\sqrt{3}\sin 20$$$$\Rightarrow \cos 20+\frac{\sin 20}{\tan x}=\cos 20+\sqrt{3}\sin 20$$$$\Rightarrow \frac{1}{\tan x}=\sqrt{3}$$$$\Rightarrow \tan x=\frac{1}{\sqrt{3}}$$$$\Rightarrow x=30°$$

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

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

$$Method2usingonlybasicgeometry:$$$$constructAEwith\mathrm{\angle }BAE=20°$$$$AB=AE=AD=ED=EC$$$$\mathrm{\angle }ECD=\mathrm{\angle }EDC=70°$$$$\Rightarrow x=70°-40°=30°$$

Commented by tanmay last updated on 21/Apr/19

$$thsnkyousir...$$

Commented by salahahmed last updated on 22/Apr/19

$$\mathrm{Why}\mathrm{is}ED=EC?$$

Commented by mr W last updated on 22/Apr/19

$$\mathrm{\angle }CAE=\mathrm{\angle }ACE=40°$$$$\Rightarrow EC=EA=ED$$

Commented by salahahmed last updated on 23/Apr/19

Thank you sir

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