Question-148630

Question Number 148630 by mathdanisur last updated on 29/Jul/21

Answered by mr W last updated on 30/Jul/21

∠C=60−20=40° ∠A=180−20−60=100° say AB=a ((AF)/(sin 20))=((BF)/(sin 100))=((AB)/(sin 60))=((2a)/( (√3))) ⇒AF=((2sin 20)/( (√3)))a ⇒BF=((2sin 100)/( (√3)))a AF+BF=(sin 20+sin 100)((2a)/( (√3)))=12 ⇒a=AB=((6(√3))/(sin 20+sin 100)) ((BC)/(sin 100))=((AB)/(sin 40)) ⇒BC=((6(√3) sin 100)/(sin 40(sin 20+sin 100))) =((6(√3))/(sin 40(2sin 10+1)))=12

$$\angle{C}=\mathrm{60}−\mathrm{20}=\mathrm{40}° \\ $$$$\angle{A}=\mathrm{180}−\mathrm{20}−\mathrm{60}=\mathrm{100}° \\ $$$${say}\:{AB}={a} \\ $$$$\frac{{AF}}{\mathrm{sin}\:\mathrm{20}}=\frac{{BF}}{\mathrm{sin}\:\mathrm{100}}=\frac{{AB}}{\mathrm{sin}\:\mathrm{60}}=\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow{AF}=\frac{\mathrm{2sin}\:\mathrm{20}}{\:\sqrt{\mathrm{3}}}{a} \\ $$$$\Rightarrow{BF}=\frac{\mathrm{2sin}\:\mathrm{100}}{\:\sqrt{\mathrm{3}}}{a} \\ $$$${AF}+{BF}=\left(\mathrm{sin}\:\mathrm{20}+\mathrm{sin}\:\mathrm{100}\right)\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}}=\mathrm{12} \\ $$$$\Rightarrow{a}={AB}=\frac{\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{sin}\:\mathrm{20}+\mathrm{sin}\:\mathrm{100}} \\ $$$$\frac{{BC}}{\mathrm{sin}\:\mathrm{100}}=\frac{{AB}}{\mathrm{sin}\:\mathrm{40}} \\ $$$$\Rightarrow{BC}=\frac{\mathrm{6}\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{100}}{\mathrm{sin}\:\mathrm{40}\left(\mathrm{sin}\:\mathrm{20}+\mathrm{sin}\:\mathrm{100}\right)} \\ $$$$\:\:\:\:=\frac{\mathrm{6}\sqrt{\mathrm{3}}}{\mathrm{sin}\:\mathrm{40}\left(\mathrm{2sin}\:\mathrm{10}+\mathrm{1}\right)}=\mathrm{12} \\ $$

Commented by mathdanisur last updated on 30/Jul/21