How do you find the height of a triangle given $3$ angles and the base side? Image given. 
This question has me absolutely stumped. This is the image of the question, how can I work out $x$? I've been doing a variety of attempts but I just cant get it.
 A: Call the side opposite $33^\circ$ as $a$. Therefore we have: $$\dfrac {x}{a}=\sin 25^\circ$$
and from the sine rule for the triangle we know that: $$\dfrac {20}{\sin 122^\circ}=\dfrac {a}{\sin 33^\circ}$$
Therefore from the above two equations we have $x=\dfrac{20\times\sin 33^\circ \times \sin 25^\circ}{\sin 122^\circ} $, or $$x\approx5.428336828982414$$
A: Another approach is to note that
$$
x\cot(33^\circ)+x\cot(25^\circ)=20
$$
to get
$$
\begin{align}
x
&=\frac{20}{\cot(33^\circ)+\cot(25^\circ)}\\[4pt]
&\approx5.42833368289824
\end{align}
$$
A: Simple, since sine rule, so $$\frac{a}{\sin(A)}=\frac{b}{\sin(B)}=\frac{c}{\sin(C)}$$ and the diagonal opposing 33° can be found by $$a=\frac{20\cdot \sin33}{\sin122}=12.844...$$ then, we use sine to find the height $$\frac{h}{12.844...}=\sin25$$and boom, $$h=12.844...\cdot \sin25=5.428...$$
A: problem statement, with construction
I prefer to think of it this way:
The base of the overall triangle can be composed of segments S and T, where:
20 = S + T

Then, using tan(A) = opposite side/adjacent side:
tan(25) =X/S
tan(33) =X/T

X is a common side, so, X = X
S*tan(25)  = T * tan(33)

But
T = B - S

so the rest is just algebra
S*tan(25) = (B-S)*tan(33)
S*tan(25) = B*tan(33) - S*tan(33)
S*tan(25) + S*tan(33) = B*tan(33)
S(tan(25) + tan(33)) = B*tan(33)
S = B * (tan(33)/(tan(25) + tan(33))

S = (20 * .649407)/(.466307+.649407)
S = 11.641099

Going back to tan(25) = X/S
X = S * tan(25)
X = 11.641099 * .466307
X = 5.42833

