Menu Close

Question-188339




Question Number 188339 by normans last updated on 28/Feb/23
Answered by universe last updated on 28/Feb/23
ar(△ABC) = ar(ABD)  +  ar(ADC)  AB.ACsin (α+β) = AB.ADsin(α) + AD.ACsin(β)  ((sin(α+β) )/(AD)) = ((sinα )/(AC)) + ((sinβ )/(AB))
$${ar}\left(\bigtriangleup{ABC}\right)\:=\:{ar}\left({ABD}\right)\:\:+\:\:{ar}\left({ADC}\right) \\ $$$${AB}.{AC}\mathrm{sin}\:\left(\alpha+\beta\right)\:=\:{AB}.{AD}\mathrm{sin}\left(\alpha\right)\:+\:{AD}.{AC}\mathrm{sin}\left(\beta\right) \\ $$$$\frac{\mathrm{sin}\left(\alpha+\beta\right)\:}{{AD}}\:=\:\frac{\mathrm{sin}\alpha\:}{{AC}}\:+\:\frac{\mathrm{sin}\beta\:}{{AB}}\: \\ $$$$\:\: \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *