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Question Number 58422 by peter frank last updated on 22/Apr/19
Answered by 2pac last updated on 23/Apr/19
$${DB}=\frac{{y}}{{sin}\left(\beta\right)} \\ $$$$<{ADB}=\mathrm{180}−\alpha−\beta \\ $$$${we}\:{have}\:\frac{{DB}}{{sin}\left(\alpha\right)}=\frac{{AB}}{{sin}\left(<{ADB}\right)}==>\frac{{y}}{{sin}\left(\alpha\right)}=\frac{{AB}}{{sin}\left(\mathrm{180}−\alpha−\beta\right)} \\ $$$${sin}\left(\mathrm{180}−{z}\right)={sinz}\:{so}\: \\ $$$${AB}=\frac{{y}}{{sin}\left(\alpha\right)}×{sin}\left(\alpha+\beta\right) \\ $$$$ \\ $$