Solving side lengths using trigonometric functions

Question

I have to find the side length of $h$ in the figure below using trigonometric functions. The whole triangle is a right triangle (I forgot to denote that in the drawing).

I know the answer is $446$ units, but I don't know how to solve this. My previous attempts ended up getting me answers that don't make sense (a hypotenuse less than the other side lengths, for example).
Can someone give a good hint or explain the solution?

Answer 1

Hint: Express $\tan(49^\circ10')$ and $\tan(29^\circ50')$ as ratios of sides. You get two equations with three unknowns. Is there a final way to relate some of the unknowns so that you get three equations?

Answer 2

Solve the following equation: $$(h\cot49^{\circ}10'+392)\tan29^{\circ}50'=h$$

Answer 3

$\tan(29^{\circ}50')=0.57=\frac{h}{392+x}$

$\tan(49^{\circ}10')=1.157=\frac{h}{x}$

These are 2 equations in two unknowns, can you take it from here?

Answer 4

Please use Law of sines to find the length of the sides.

In particular, denote the line in the middle by $a$.

Then from triangle on the left you know that $$\frac{a}{\sin 29^{\circ} 50'} = \frac{392}{\sin 19^{\circ}20'} $$

Note, the second angle of $19^{\circ}20'$, it comes from the the external angle equation, i.e. $\alpha + 29^{\circ}50' = 49^{\circ}10'$.

From triangle on the right $$\frac{a}{\sin 90^{\circ}} = \frac{h}{\sin 49^{\circ}10'} $$

Therefore, $$h = 392 \cdot \frac{\sin 29^{\circ} 50'}{\sin 19^{\circ}20'} \cdot \sin 49^{\circ}10' \approx 446 $$