Solving problem with Sines Law and Compound Angle Formula

X  is any point on the side BC of a triangle ABC . If the $ \displaystyle \angle AXC=\theta \ and\ \frac{{BX}}{{XC}}=\frac{m}{n}$ , show that $ \displaystyle (m+n)\cos \theta =n\cot B-m\cot C$ .

Solution

$ \displaystyle \text{Proof}:\ \ \text{By}\ \text{the}\ \text{law}\ \text{of}\ \text{sines}\ ,$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \frac{{BX}}{{\sin \beta }}=\frac{{AB}}{{\sin ({{{180}}^{\circ }}-\theta )}}=\frac{{AB}}{{\sin \theta }}\ \ .....(1)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \frac{{XC}}{{\sin \alpha }}=\frac{{AC}}{{\sin \theta }}\ \ \ ..........(2)$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \frac{{AB}}{{\sin C}}=\frac{{AC}}{{\sin B}}\ \ \ ..........(3)$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ (1)\div (2)\Rightarrow \ \ \ \frac{{BX\sin \alpha }}{{XC\sin \beta }}=\frac{{AB}}{{AC}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{m\sin \alpha }}{{n\sin \beta }}=\frac{{\sin C}}{{\sin B}}\ \ \ \ [\text{By}\ (3)]$ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{m\sin ({{{180}}^{\circ }}-(\theta +C))}}{{n\sin (\theta -B)}}=\frac{{\sin C}}{{\sin B}}\ $

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{{m\sin (\theta +C)}}{{n\sin (\theta -B)}}=\frac{{\sin C}}{{\sin B}}\ $ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \frac{{m(\sin \theta \cos C+\cos \theta \sin C)}}{{\sin C}}=\frac{{n(\sin \theta \cos B-\cos \theta \sin B)}}{{\sin B}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ m\sin \theta \cot C+m\cos \theta =n\sin \theta \cot B-n\cos \theta $ 

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (m+n)\cos \theta \ =\sin \theta \ (\ n\cot B-m\cot C\ )$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \therefore \ \ \ \ (m+n)\cot \theta \ =\ n\cot B-m\cot C$