trigonometry
I'm having issues with the following problem, which I think are due to how I am modelling the problem. Seems like I have to think in 2D and 3D which is where I get confused. Could someone show me how they would approach modelling this?
Question:
From a point A the bearing of the base of a tower 60m high is 53° and the angle of elevation of the top of the tower is 16°. From a point B the bearing is 300° and the angle of elevation is 20°. Find the distance AB.
BF is the ground level.
Write EG in terms of angle $50^\circ$.Find $x$. Then the height will be $x\tan 46^\circ + 10$
Let $h$ be the unknown height of the tower. Let $d$ be the (equally unknown) distance from $P$, the point we first set up our equipment, to the base $B$ of the tower. Let $\theta$ be the measured angle of elevation. So in your case, $\theta$ is $26^\circ 50'$.
Draw the right-angled triangle $TBP$, where $T$ is the top of the tower. Then $$\frac{h}{d}=\tan\theta.$$
Now repeat, letting $Q$ be the second point of measurement. The distance from $Q$ to the base of the tower is $d-25$. Let $\phi$ be the angle of elevation from $Q$. You were given $\phi$.
By the same reasoning as before, we have $$\frac{h}{d-25}=\tan\phi.$$
We have two equations, in the two unknowns $h$ and $d$. We want to solve for $h$.
Take the reciprocal of each side of the first equation. We get $$\frac{d}{h}=\frac{1}{\tan \theta}.\tag{$1$}$$
Similarly, take the reciprocal of each side of the second equation. We get $$\frac{d-25}{h}=\frac{d}{h}-\frac{25}{h}=\frac{1}{\tan \phi}.\tag{$1$}$$ Look at the two equations $(1)$ and $(2)$. Subtract each side of $(2)$ from the corresponding side of $(1)$. We get $$\frac{25}{h}=\frac{1}{\tan\theta}-\frac{1}{\tan\phi}.$$ Now solving for $h$ is just some algebra. We get $$h=25\left(\frac{\tan\theta\tan\phi}{\tan\phi-\tan\theta} \right).\tag{$3$}$$
Remark: You asked about attempting what I would call a linear approximation. Good idea. I will not go through the details of how one might try to do it that way. The problem is that in surveying, we try to achieve very high standards of accuracy. A rough estimate is not good enough. And we don't have to settle for a rough estimate, since, at least to the standards of prcision of our measurements, we can get an "exact" answer from Formula $(3)$.
Best Answer
What does the elevation tell you? Let's say the base of the tower is $C$, the top is $D$. From point $A$, assumed to be at the same height as the base of the tower, the tangent of the elevation angle gives you the ratio $\frac{DC}{AC}$. You know $DC=60$, so you can find $AC$. Similarly, find the length of the $BC$ line. Now we draw the $ABC$ triangle. Let's choose the tower at origin. We know that from $A$ the bearing is $53^\circ$. That means that from $C$ you see $A$ at $53^\circ\pm 180^\circ$. Using a similar reasoning, the $C$ point is at the $300^\circ\pm 180^\circ$. Now you know $AC$, $BC$, and the angle $BCA$, so just apply the generalized Pythagoras's theorem to find $AB$