trigonometry
"The famous leaning tower of pisa 'leans' at an angle of 4 degrees, and its top is 56.86 meters above the ground. Determine the height of its top above the ground when it was originally built (standing at right angles to the ground."
Using a quickly drawn illustration, I came to the answer that the original height was also 55.86, however, this seems wrong in my eyes, as the leaning tower's height from the ground is then the same length.
I hope someone can point out where I went wrong. Thanks!
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 $(3)$.
Consider the following figure.
Points $B$ and $C$ are the endpoints of the bridge. Point $A$ is the point on the ground from which the angles of elevation are measured.
Observe that
\begin{align*} \tan(68^\circ) & = \frac{h}{x}\\ \tan(62^\circ) & = \frac{h}{10~\text{m} - x} \end{align*}
Solving the equation
$$\tan(68^\circ) = \frac{h}{x}$$
for $h$ yields
$$h = x\tan(68^\circ)$$
Solving the equation
$$\tan(62^\circ) = \frac{h}{10~\text{m} - x}$$
for $h$ yields
$$h = (10~\text{m} - x)\tan(62^\circ)$$
We can solve for $x$ by equating the two expressions for $h$.
\begin{align*} x\tan(68^\circ) & = (10~\text{m} - x)\tan(62^\circ)\\ x\tan(68^\circ) & = 10~\text{m}\tan(62^\circ) - x\tan(62^\circ)\\ x\tan(68^\circ) + x\tan(62^\circ) & = 10~\text{m}\tan(62^\circ)\\ x[\tan(68^\circ) + \tan(62^\circ)] & = 10~\text{m}\tan(62^\circ)\\ x & = \frac{10~\text{m}\tan(62^\circ)}{\tan(68^\circ) + \tan(62^\circ)} \end{align*}
Substituting for $x$ in the equation
$$h = x\tan(68^\circ)$$
yields
$$h = \frac{10~\text{m}\tan(62^\circ)\tan(68^\circ)}{\tan(68^\circ) + \tan(62^\circ)} \approx 10.69~\text{m}$$
which is about the height of a three-story building.
Best Answer
Clearly the original height would've been greater than the vertical height after slanting.
Draw a right angled triangle. The length of tower measured along its side (which is its original height) is the hypotenuse $H$. The lean angle is to the vertical (it would be absurd to think it leans $4$ degrees to the horizontal). That means you have the relationship $\cos 4^{\circ} = \frac{56.86}{H}$ from which you quickly get $H = 56.9988$ metres.
Excuse the crappy diagram.