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In triangle ABC, \(\angle B = 60^\circ\) and \(\angle C = 45^\circ.\) The point D divides BC in the ratio 1:3. Find \(\frac{\sin \angle BAD}{\sin \angle CAD}.\)
 

 Oct 4, 2019
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By the Law of Sines

 

sin BAD / BD  = sin 60/AD                sin CAD / DC  = sin 45 / AD

sin BAD / 1  = sin 60 / AD       and    sin  CAD / 3  = sin 45 / AD

sin BAD = sin 60 / AD                       sin CAD  =  3 sin 45  / AD

 

So

 

sin BAD            sin 60 / AD           √3/2              √3/2                                          

_______  =     ___________  =    _______  =   _____  =   √3/2  * √2/3  = √6/6 = 1/√6  

sin CAD            3 sin 45 / AD         3 * 1/√2         3/√2

 

cool cool cool

 Oct 5, 2019
edited by CPhill  Oct 5, 2019
edited by CPhill  Oct 5, 2019

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