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Question-206729




Question Number 206729 by mr W last updated on 23/Apr/24
Answered by A5T last updated on 23/Apr/24
((sin84)/(AD))=((sin54)/(AC))⇒AD=((ACsin84)/(sin54))...(i)  ((sin(?))/(AD))=((sin(54−?))/(BD=AC))⇒AD=((ACsin(?))/(sin(54−?)))...(ii)  (i)&(ii)⇒((sin84)/(sin54))=((sin(?))/(sin(54−?)))⇒?=30°
$$\frac{{sin}\mathrm{84}}{{AD}}=\frac{{sin}\mathrm{54}}{{AC}}\Rightarrow{AD}=\frac{{ACsin}\mathrm{84}}{{sin}\mathrm{54}}…\left({i}\right) \\ $$$$\frac{{sin}\left(?\right)}{{AD}}=\frac{{sin}\left(\mathrm{54}−?\right)}{{BD}={AC}}\Rightarrow{AD}=\frac{{ACsin}\left(?\right)}{{sin}\left(\mathrm{54}−?\right)}…\left({ii}\right) \\ $$$$\left({i}\right)\&\left({ii}\right)\Rightarrow\frac{{sin}\mathrm{84}}{{sin}\mathrm{54}}=\frac{{sin}\left(?\right)}{{sin}\left(\mathrm{54}−?\right)}\Rightarrow?=\mathrm{30}° \\ $$
Commented by mr W last updated on 23/Apr/24
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