# Thread: tower question

1. ## tower question

China's leaning tower has a height of 47m. It makes an angle of 87.2 degrees with the ground. There is a specific point on the ground where you can be equidistant from both the top and bottom of the tower. How far is this point from the base of the tower?

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