1. ## trigo (5)

The point D divides the side AB of the triangle ABC internally in the ratio of m:n . If angle ACD =a , angle BCD = b , and angle BDC = \theta , use the sine rule to show taht

$\displaystyle \frac{AD}{sin a}=\frac{DC}{sin(\theta-a)}$

$\displaystyle \frac{BD}{sin b }=\frac{DC}{sin(\theta+b)}$

I can show this part . The one below is what i am not sure of .

Hence , prove that $\displaystyle (m+n)\cot \theta=m\cot a-n\cos b$

2. Originally Posted by thereddevils
The point D divides the side AB of the triangle ABC internally in the ratio of m:n . If angle ACD =a , angle BCD = b , and angle BDC = \theta , use the sine rule to show that

$\displaystyle \frac{AD}{sin a}=\frac{DC}{sin(\theta-a)}$

$\displaystyle \frac{BD}{sin b }=\frac{DC}{sin(\theta+b)}$

I can show this part . The one below is what i am not sure of .

Hence , prove that $\displaystyle (m+n)\cot \theta=m\cot a-n\,{\color{red}\cot}\, b$
You are told that AD/BD = m/n. Plug the values $\displaystyle AD = \frac{DC\sin a}{\sin(\theta-a)}$ and $\displaystyle BD = \frac{DC\sin b}{\sin(\theta+b)}$ into that equation, and you get $\displaystyle \frac{\sin a\sin(\theta+b)}{\sin b\sin(\theta-a)} = \frac mn$. Multiply out the fractions, use the addition formulae for the sine function, and it becomes $\displaystyle n\sin a(\sin\theta\cos b + \cos\theta \sin b) = m\sin b(\sin\theta\cos a - \cos\theta\sin a)$. Now divide by $\displaystyle \sin a\sin b\sin\theta$ to get $\displaystyle n(\cot b + \cot\theta) = m(\cot a - \cot\theta)$, and you're nearly there... .