# Find a side of a triangle given other two sides and an angle

I have a really simple-looking question, but I have no clue how I can go about solving it?

The question is

Find the exact value of $x$ in the following diagram:

Sorry for the silly/easy question, but I'm quite stuck! To use the cosine or sine rule, I'd need the angle opposite $x$, but I can't find that, cause I don't have anything else to help it along.

Thank You!

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Call the top point $A$, the point on the bottom left $B$, and the point on the bottom right $C$. Draw the altitude from $A$ to $BC$, and call the foot of the altitude $D$.

Now $\triangle ABD$ has angles $30^\circ$, $60^\circ$, and $90^\circ$ respectively. Therefore $BD$ has length $3$ and $AD$ has length $3\sqrt3$. Furthermore, by the Pythagorean theorem, $DC$ has length $\sqrt{7^2 - (3\sqrt3)^2} = \sqrt{22}$.

Therefore $x = BC = BD + DC = 3 + \sqrt{22}$.

(Forgive my shoddy length notation.)

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Use the cosine rule with respect to the 60 degree angle. Then you get an equation involving $x$ as a variable, Then you solve the equation for $x$.

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Good hint. Better than the one I was typing up. –  Ross Millikan Aug 18 '12 at 4:26
$$\rm{Cos}\,60^{\circ}=\frac{6^2+x^2-7^2}{2\times6\times x},$$
Then you can easily find out that $x=3+\sqrt{22}$.