Hello, andrewpark2!

China's leaning tower has a height of 47m and makes an angle of 87.2° with the ground.

There is a point on the ground which is equidistant from the top and bottom of the tower.

How far is this point from the base of the tower? Code:

A
o
*
* 87.2°*
x *
* * 47
*
* *
* 5.6° 87.2°
C o * * * * * * o B
x

The tower is: $\displaystyle AB = 47$

$\displaystyle \angle B = 87.2^o$

The special point is at $\displaystyle C\!:\;\;x = CA =CB $

Since $\displaystyle \Delta ABC$ is isosceles: .$\displaystyle \angle A = 87.2^o,\;\angle C = 5.6^o$

Law of Cosines: .$\displaystyle 47^2 \;=\;x^2+x^2-2(x)(x)\cos5.6^o$

. . . . . $\displaystyle 47^2 \:=\:2x^2 - 2x^2\cos5.6^o \;=\;2x^2(1 - \cos5.6^o) \;=\;4x^2\left(\frac{1-\cos5.6^o}{2}\right)$

. . . . . $\displaystyle 47^2 \;=\;4x^2\sin^2\!2.8^o \quad\Rightarrow\quad x^2 \:=\:\frac{47^2}{4\sin^2\!2.8^o} \quad\Rightarrow\quad x \:=\:\frac{47}{2\sin2.8^o}$

Therefore: .$\displaystyle x \;=\;481.0667... \;\approx\;481\text{ m}$