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Sepia's question at Yahoo! Answers regarding finding the length of a line segment within a triangle

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MarkFL

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Feb 24, 2012
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Here is the question:

How will you find the length of a line drawn inside a scalene triangle?


ABC is a scalene triangle, in which AB = 3, AC = 4 and BC = 6
D is the line joining A with BC.such that BD : DC : :2 :3
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Re: Sepia's question ay Yahoo! Answers regarding finding the length of a line segment within a trian

Hello Sepia,

Let's first draw a diagram:

sepia.jpg

We see that:

\(\displaystyle u+v=6\)

and we are given:

\(\displaystyle \frac{u}{v}=\frac{2}{3}\implies u=\frac{2}{3}v\)

Substituting this into the first equation, we then find:

\(\displaystyle \frac{2}{3}v+v=6\)

\(\displaystyle \frac{5}{3}v=6\)

\(\displaystyle v=\frac{18}{5}\)

Now, let's use the Law of Cosines to determine the cosine of the angle $\theta$:

\(\displaystyle 3^2=4^2+6^2-2\cdot4\cdot6\cos(\theta)\)

\(\displaystyle \cos(\theta)=\frac{4^2+6^2-3^2}{2\cdot4\cdot6}=\frac{43}{48}\)

Next, using the Law of Cosines again, we may state:

\(\displaystyle x=\sqrt{4^2+v^2-2\cdot4\cdot v\cos(\theta)}\)

\(\displaystyle x=\sqrt{4^2+\left(\frac{18}{5} \right)^2-2\cdot4\cdot\left(\frac{18}{5} \right)\left(\frac{43}{48} \right)}=\frac{\sqrt{79}}{5}\)