How do you find the height of a triangle given $3$ angles and the base side? Image given.

Question

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.

Answer 1

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 $$x\approx5.428336828982414$$

Answer 2

Another approach is to note that $$ x\cot(33^\circ)+x\cot(25^\circ)=20 $$ to get $$ \begin{align} x &=\frac{20}{\cot(33^\circ)+\cot(25^\circ)}\\[4pt] &\approx5.42833368289824 \end{align} $$