Hello, fayeorwhatsoever!

1.) From a point 55 m up the ground on a vertical antenna pole, three cables are stretched

into anchors which are located at the vertices of an equilateral triangle 20 m on a side.

Find the length of each cable if the foot of the pole is equidistant from the anchors.

Looking down at the ground, the anchors are situated like this: Code:

A
*
/|\
/ | \
/ | \
20 / | \ 20
/ | \
/ | \
/ *G \
/ | \
/ | \
B * - - - - * - - - - * C
10 D 10

The pole is at $\displaystyle G$, which is equidistant from the vertices.

This point is at the centroid of the triangle,

. . which divides the altitude in the ratio 2:1.

Using Pythagorus, we find that the altitude is: .$\displaystyle AD \:=\:10\sqrt{3}$

Then G is located so that: .$\displaystyle AG \:=\:\frac{2}{3}(10\sqrt{3}) \:=\:\frac{20\sqrt{3}}{3}$

A side view of the pole and cable looks like this: Code:

F *
|\
| \
| \
| \
55 | \
| \
| \
| \
| \
G * - - - - * A
(20√3)/3

The pole is: $\displaystyle FG \,=\,55$

The anchor is at $\displaystyle A\!:\;\;GA \,=\,\frac{20\sqrt{3}}{3}\,=\,\frac{20}{\sqrt{3}}$

The length of the cable is the hypotenuse:

. . $\displaystyle FA^2 \;=\;55^2 + \left(\frac{20}{\sqrt{3}}\right)^2 \;=\; 3025 + \frac{400}{3} \;=\;\frac{9475}{3}$

Therefore: .$\displaystyle FA \;=\;\sqrt{\frac{9475}{3}} \;=\;\frac{5}{3}\sqrt{1137} \;\approx\;56.2\text{ m}$