Question Number 193203 by Mingma last updated on 07/Jun/23
Answered by som(math1967) last updated on 07/Jun/23
![((DB)/(DC))=((BE)/(EC)) [DE is bisector of∠BDC] ((DB)/(DC))=(1/2) Sin∠DCB=Sin30 ∠DCB=30 ((DB)/(BC))=tan30 DB=(√3) [BC=3] ((DB)/x)=((BE)/(EC)) [DE∥AC] ∴ x=2(√3) ans AB=2(√3)+(√3)=(√3) tan∠DAC=((BC)/(AB))=(3/(3(√3)))=(1/( (√3)))=tan30 ∴∠DAC=30](https://www.tinkutara.com/question/Q193206.png)
$$\frac{{DB}}{{DC}}=\frac{{BE}}{{EC}}\:\:\left[{DE}\:{is}\:{bisector}\:{of}\angle{BDC}\right] \\ $$$$\:\frac{{DB}}{{DC}}=\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$${Sin}\angle{DCB}={Sin}\mathrm{30} \\ $$$$\angle{DCB}=\mathrm{30} \\ $$$$\:\frac{{DB}}{{BC}}={tan}\mathrm{30} \\ $$$${DB}=\sqrt{\mathrm{3}}\:\:\:\left[{BC}=\mathrm{3}\right] \\ $$$$\frac{{DB}}{{x}}=\frac{{BE}}{{EC}}\:\:\:\left[{DE}\parallel{AC}\right] \\ $$$$\:\therefore\:{x}=\mathrm{2}\sqrt{\mathrm{3}}\:\:\:\:\:\:\:{ans} \\ $$$$\:{AB}=\mathrm{2}\sqrt{\mathrm{3}}+\sqrt{\mathrm{3}}=\sqrt{\mathrm{3}} \\ $$$$\:{tan}\angle{DAC}=\frac{{BC}}{{AB}}=\frac{\mathrm{3}}{\mathrm{3}\sqrt{\mathrm{3}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}={tan}\mathrm{30} \\ $$$$\therefore\angle{DAC}=\mathrm{30} \\ $$$$ \\ $$
Commented by Mingma last updated on 07/Jun/23
Perfect ��