trigo (5)

**thereddevils** · August 14th 2009, 04:30 AM

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

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

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

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

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

**Opalg** · August 14th 2009, 05:03 AM

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

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

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

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

Hence , prove that $(m+n)\cot \theta=m\cot a-n\,{\color{red}\cot}\, b$

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

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