Question

Sollutions of triangles 2

Answer 1

In a triangle ABC, prove that for any \[\huge \angle \theta \]
\[\huge bcos(A-\theta) +acos(B+\theta) = ccos \theta \]

Answer 2

@sidsiddhartha , @Owlfred , @Miracrown

Answer 3

Divide both sides by \(\cos \theta\).

Answer 4

\[\dfrac{\cos(A - \theta)}{\cos\theta}= \dfrac{\cos A \cos \theta + \sin A \sin \theta }{\cos\theta}=\cos A + \sin A \tan\theta\]\[\dfrac{\cos( B + \theta)}{\cos \theta }=\dfrac{\cos B \cos \theta - \sin B \sin \theta}{\cos\theta}=\cos B - \sin B \tan \theta\]So we have...\[b (\cos A + \sin A \tan \theta) + a (\cos B - \sin B \tan \theta) = c\]\[a \cos B + b\cos A +\tan \theta \left(b \sin A -a \sin B\right )=c\]Now by sine rule, \(b\sin A - a\sin B = 0\).
Therefore we are left with\[a \cos B + b \cos A = c ~\checkmark \]

Answer 5

Thank you , I appreciate your help