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Question-204560




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 ∠B=?
$${AC}={BD} \\ $$$${find}\:\angle{B}=? \\ $$
Answered by es last updated on 21/Feb/24
AC=BD  ((AC)/(sin80))=((AD)/(sin40))  →AD=((AC)/(2cos40))  ((AD)/(sin?))=((BD)/(sin(100−?)))→  ((AC)/(2cos40×sin?))=((BD)/(sin(100−?)))→  ?=41.53
$${AC}={BD} \\ $$$$\frac{{AC}}{{sin}\mathrm{80}}=\frac{{AD}}{{sin}\mathrm{40}}\:\:\rightarrow{AD}=\frac{{AC}}{\mathrm{2}{cos}\mathrm{40}} \\ $$$$\frac{{AD}}{{sin}?}=\frac{{BD}}{{sin}\left(\mathrm{100}−?\right)}\rightarrow \\ $$$$\frac{{AC}}{\mathrm{2}{cos}\mathrm{40}×{sin}?}=\frac{{BD}}{{sin}\left(\mathrm{100}−?\right)}\rightarrow \\ $$$$?=\mathrm{41}.\mathrm{53} \\ $$$$ \\ $$$$ \\ $$
Commented by A5T last updated on 21/Feb/24
((AD)/(sin ?))=((BD)/(sin(80−?)))⇒((AC)/(2cos40sin ?))=((AC)/(sin(80−?)))  ⇒?=30°
$$\frac{{AD}}{{sin}\:?}=\frac{{BD}}{{sin}\left(\mathrm{80}−?\right)}\Rightarrow\frac{{AC}}{\mathrm{2}{cos}\mathrm{40}{sin}\:?}=\frac{{AC}}{{sin}\left(\mathrm{80}−?\right)} \\ $$$$\Rightarrow?=\mathrm{30}° \\ $$
Commented by es last updated on 22/Feb/24
thanks   ⋛
$${thanks}\:\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Answered by Ghisom last updated on 21/Feb/24
((AC)/(sin 80°))=((AD)/(sin 40°))  ((AC)/(sin (80°−β)))=((AD)/(sin β))  −−−−−−−−−  this transforms to  tan β =((√3)/3)  β=30°
$$\frac{{AC}}{\mathrm{sin}\:\mathrm{80}°}=\frac{{AD}}{\mathrm{sin}\:\mathrm{40}°} \\ $$$$\frac{{AC}}{\mathrm{sin}\:\left(\mathrm{80}°−\beta\right)}=\frac{{AD}}{\mathrm{sin}\:\beta} \\ $$$$−−−−−−−−− \\ $$$$\mathrm{this}\:\mathrm{transforms}\:\mathrm{to} \\ $$$$\mathrm{tan}\:\beta\:=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\beta=\mathrm{30}° \\ $$

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