Hello, tiar!

A flagpole is located at the edge of a sheer 50ft cliff at the bank of a river 40 feet wide.

An observer on the opposite side of the river measures an angle of 9 degrees between

her line of sight to the top of the flagpole and her line of sight to the top of the cliff.

Find the height of the flagpole. Code:

o A
* |
* |
* | h
* |
* o B
* * |
* 9° * |
* * | 50
* * θ |
P o - - - - - - - - - o C
40

The flagpole is $\displaystyle h = AB.$

The cliff is $\displaystyle BC = 50.$

The observer is at $\displaystyle P\!:\;\;PC = 40.$

$\displaystyle \angle APB = 9^o.$

Let $\displaystyle \theta = \angle BPC.$

In right triangle $\displaystyle BCP\!:\;\tan\theta = \tfrac{50}{40} = \tfrac{5}{4}$

In right triangle $\displaystyle ACP\!:\;\;\tan(\theta + 9^o) \:=\:\frac{h+50}{40}\;\;{\color{blue}[1]}$

. . Note that: .$\displaystyle \tan(\theta + 9^o) \:=\:\frac{\tan\theta + \tan9^o}{1-\tan\theta\tan9^o} \;=\;\frac{\frac{5}{4} + \tan9^o}{1 - \frac{5}{4}\tan9^o}$ .$\displaystyle = \:\frac{5+4\tan9^o}{4-5\tan9^o} $

Substitute into [1]: .$\displaystyle \frac{5+\tan9^o}{4-5\tan9^o} \;=\;\frac{h+50}{40} \quad\Rightarrow\quad h \;=\;40\cdot\frac{5 + 4\tan9^o}{4-5\tan9^o} - 50 $

Therefore: .$\displaystyle h \;=\;20.24190952 \;\approx\;20.24\text{ ft}$