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Question Number 193203 by Mingma last updated on 07/Jun/23

Answered by som(math1967) last updated on 07/Jun/23

$$\frac{DB}{DC}=\frac{BE}{EC}\left[DEisbisectorof\mathrm{\angle }BDC\right]$$$$\frac{DB}{DC}=\frac{1}{2}$$$$Sin\mathrm{\angle }DCB=Sin30$$$$\mathrm{\angle }DCB=30$$$$\frac{DB}{BC}=tan30$$$$DB=\sqrt{3}[BC=3]$$$$\frac{DB}{x}=\frac{BE}{EC}[DE\parallel AC]$$$$\therefore x=2\sqrt{3}ans$$$$AB=2\sqrt{3}+\sqrt{3}=\sqrt{3}$$$$tan\mathrm{\angle }DAC=\frac{BC}{AB}=\frac{3}{3\sqrt{3}}=\frac{1}{\sqrt{3}}=tan30$$$$\therefore \mathrm{\angle }DAC=30$$$$$$

Commented by Mingma last updated on 07/Jun/23

Perfect ��

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