A triangle with angles measuring $15^\circ$, $45^\circ$, and $120^\circ$. The side opposite the $45^\circ$ angle is $20$ units in length and the area of the triangle in square units as $(m-n\sqrt{q})$, where $q$ is a prime number. What is the value of the sum $m+n+q$?
I have thought a lot about this problem, I thought about extending the side opposite the $15^\circ$ angle to get a 45-45-90 triangle. From this, we could subtract an area from the area of the overall triangle to get the area we want to get, but I haven't been able to do this yet. Can anyone help, please?
Thanks!
I tried using trig at one point, but I'm probably making a mistake since that just gave me an answer with a bunch of sin and cos expressions.
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$\begingroup$I have provided an image to show you this as I thought it would provide the clearest solution.
I have used the sine rule (or law of sines) to find one of the sides and used the formula area$=\frac{1}{2}ac\sin B$.
I used a calculator here, but you could easily calculate $\sin 15$ using the compound angle formulae if this was from some competition. Hope this helps.
Let $h$ be the altitude drawn from the $45$-degree vertex. Then,
$$20= h\cot 15 + h \cot120$$
and the area is
$$ A=\frac12 \cdot 20 h = \frac{200}{\cot 15 + \cot120} = \frac{200}{\cot (45-30) - \cot60} =50(3-\sqrt3).$$
Thus, $ m+n+q =203$.
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