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I need help with this exercise I got.

We have a triangle with given medians ma=6, mb=9 and side (without given median on that side) c=6. What is the length of a and b and with what values of ma,mb and c we wouldn't have a solution?

I've came to the solution (using trigonometry) that a is around 6 and b 0,7 but I'm not sure of it.

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What is your method. We can use… – lab bhattacharjee Mar 30 '14 at 10:10
I used cosine rule. – Breldor Mar 30 '14 at 10:13
up vote 1 down vote accepted

Using these formulas which lab bhattacharjee already pointed out in a comment, you get

$$ 2b^2+2c^2-a^2 = 4m_a^2 = 144 \\ 2a^2+2c^2-b^2 = 4m_b^2 = 324 $$

Subtract $2c^2=72$ from both sides and you get

\begin{align*} 2b^2-a^2 &= 72 & 2a^2-b^2 &= 252 \\ 3a^2 &= 2\cdot 252+72 = 576 & 3b^2 &= 2\cdot72+252 = 396 \\ a^2 &= 192 & b^2 &= 132 \\ a &= \sqrt{192} = 8\sqrt3 \approx 13.8564 & b &= \sqrt{132} = 2\sqrt{33} \approx 11.4891 \end{align*}

So your triangle looks like this:


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