# Using trig functions with no known sides

Here is a question I need to work through:

"From a point P on level ground, the angle of elevation of the top of a tower is 26 degrees and 50 minutes. From a point 25.0 meters closer to the tower and on the same line with P and the base of the tower, the angle of elevation of the top is 53 degrees and 30 minutes. Approximate the height of the tower."

My question is, does the angle of elevation increase towards 90 degrees at the same rate as the distance of P from the base of tower decreases towards 0? In other words could I put (53.5-(26+50/60)/(90-(26+50/60) = 25/X and then solve for X to get the distance of P to the base? It seems like it would work but I've not learned anything about it.

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Have you drawn the picture? –  Lubin Nov 29 '12 at 21:03
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## 1 Answer

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 $(3)$.

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