solution in the table or a scrap piece of
Two tracking stations $35 \mathrm{~km}$ apart measure the angles of elevation of a rocket to be $34^{\circ}$ and $82^{\circ}$. Find the height of the rocket.
Hint: Set up two equations that both use $h$

Step 1 ：Let the height of the rocket be $h$.

Step 2 ：In the triangle formed by the rocket and the two tracking stations, let the distance from the first station to the point on the ground directly below the rocket be $x$.

Step 3 ：Use the tangent function to set up two equations: \(\tan{34^\circ} = \frac{h}{x}\) and \(\tan{82^\circ} = \frac{h}{35-x}\).

Step 4 ：Solve the first equation for $x$: \(x = \frac{h}{\tan{34^\circ}}\).

Step 5 ：Substitute this expression for $x$ in the second equation: \(\tan{82^\circ} = \frac{h}{35 - \frac{h}{\tan{34^\circ}}}\).

Step 6 ：Multiply both sides by \(35 - \frac{h}{\tan{34^\circ}}\) to eliminate the denominator: \(h\tan{82^\circ} = 35\tan{82^\circ} - h\tan{34^\circ}\tan{82^\circ}\).

Step 7 ：Add \(h\tan{34^\circ}\tan{82^\circ}\) to both sides and factor out $h$: \(h(\tan{82^\circ} + \tan{34^\circ}\tan{82^\circ}) = 35\tan{82^\circ}\).

Step 8 ：Divide both sides by \(\tan{82^\circ} + \tan{34^\circ}\tan{82^\circ}\) to solve for $h$: \(h = \frac{35\tan{82^\circ}}{\tan{82^\circ} + \tan{34^\circ}\tan{82^\circ}}\).

Step 9 ：Calculate the value of $h$: \(h \approx 34.64\).

Step 10 ：\(\boxed{34.64}\)