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Math Help - trigo (5)

  1. #1
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    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

    \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
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    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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