Question Number 204560 by mr W last updated on 21/Feb/24
Commented by mr W last updated on 21/Feb/24
$${AC}={BD} \\ $$$${find}\:\angle{B}=? \\ $$
Answered by es last updated on 21/Feb/24
$${AC}={BD} \\ $$$$\frac{{AC}}{{sin}\mathrm{80}}=\frac{{AD}}{{sin}\mathrm{40}}\:\:\rightarrow{AD}=\frac{{AC}}{\mathrm{2}{cos}\mathrm{40}} \\ $$$$\frac{{AD}}{{sin}?}=\frac{{BD}}{{sin}\left(\mathrm{100}−?\right)}\rightarrow \\ $$$$\frac{{AC}}{\mathrm{2}{cos}\mathrm{40}×{sin}?}=\frac{{BD}}{{sin}\left(\mathrm{100}−?\right)}\rightarrow \\ $$$$?=\mathrm{41}.\mathrm{53} \\ $$$$ \\ $$$$ \\ $$
Commented by A5T last updated on 21/Feb/24
$$\frac{{AD}}{{sin}\:?}=\frac{{BD}}{{sin}\left(\mathrm{80}−?\right)}\Rightarrow\frac{{AC}}{\mathrm{2}{cos}\mathrm{40}{sin}\:?}=\frac{{AC}}{{sin}\left(\mathrm{80}−?\right)} \\ $$$$\Rightarrow?=\mathrm{30}° \\ $$
Commented by es last updated on 22/Feb/24
$${thanks}\:\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Answered by Ghisom last updated on 21/Feb/24
$$\frac{{AC}}{\mathrm{sin}\:\mathrm{80}°}=\frac{{AD}}{\mathrm{sin}\:\mathrm{40}°} \\ $$$$\frac{{AC}}{\mathrm{sin}\:\left(\mathrm{80}°−\beta\right)}=\frac{{AD}}{\mathrm{sin}\:\beta} \\ $$$$−−−−−−−−− \\ $$$$\mathrm{this}\:\mathrm{transforms}\:\mathrm{to} \\ $$$$\mathrm{tan}\:\beta\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\beta=\mathrm{30}° \\ $$