Three Dimensional Trigonometry

Question

From an observation point at sea level, an aircraft is observed on a bearing of N30E, elevation 20 degrees and at height 600 m. The aircraft then flies due east for 1000 m, without altering its height. What will be its bearing and angle of elevation from the observation point then? (Give answers to nearest degree). I have actually done this question. I got the answers to be 51.02 (new bearing, which is 51 to nearest degree) and 14.39 angle of elevation (14 to nearest degree). The actual answers are 52 and 15. So I want to check whether I am making a mistake or the book made a mistake with this one question. I used basic trig to get unknown sides and then used cosine rule to determine the length of the hypotenuse of the second triangle formed when the plane goes 1000 m due east. After this I used the sine rule to determine the unknown angle in the triangle which is formed from the two hypotenuses of the two triangles plus the 1000 m which forms the top side of the triangle. This gave me the unknown angle of 21.02 degrees, which I then added to the 30 degree bearing; then angle of elevation worked out using basic trig.

Answer 1

Without much more detail it’s hard to see exactly where you went wrong, but I think you made a mistake somewhere.

We are given that the elevation is $20^\circ$ and the height is $600$m. From this, we can deduce that the original distance $d_0$ (along the ground) to the aircraft is $\frac{600}{\tan(20^\circ)} \approx 1648.486$m.

Now we wish to obtain the bearing after travelling east. First, we calculate the ‘distance north’ $d_N$ to aircraft to be $d_0 \times \cos(30^\circ) \approx 1427.631$m and the ‘distance east’ $d_E$ to the aircraft to be $d_0 \times \sin(30^\circ) \approx 824.243$m. These two distances are the legs of the right-angled triangle with the line along the ground to the aircraft being the hypotenuse. The new bearing will then be $\tan^{-1}(\frac{d_E + 1000}{d_N}) \approx \mathbf{51.953^\circ}$ as the book states.

To obtain the new elevation we can calculate the new distance $d_1$ (along the ground) to be $\sqrt{{d_N}^2 + (d_E + 1000)^2} \approx 2316.43$m. We could have used the angle we calculated earlier but I used the Pythagorean theorem to preserve precision. The elevation is then $\tan^{-1}(\frac{600}{d_1}) \approx \mathbf{14.52^\circ}$, which also matches the book’s answer.

(Note that I didn’t use any fancy rules; just basic trigonometric functions.)