# Geometry problem

• December 13th 2008, 02:19 PM
fayeorwhatsoever
Geometry problem
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.

2.) A flat roof has an inclination of 30 degrees to the floor of a building. The roof is 50 m long (being parallel to the floor) and 20 m wide. Find the area of the projection of the roof upon the floor of the building.

3.) What is the area of the projection on the plane of an equilateral triangle 10 cm on a side, if one side is parallel to the plane and the other sides form an angle of 30 degrees with the plane?

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• December 13th 2008, 06:40 PM
Soroban
Hello, fayeorwhatsoever!

Quote:

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 $G$, 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: . $AD \:=\:10\sqrt{3}$

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

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

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

The pole is: $FG \,=\,55$
The anchor is at $A\!:\;\;GA \,=\,\frac{20\sqrt{3}}{3}\,=\,\frac{20}{\sqrt{3}}$

The length of the cable is the hypotenuse:
. . $FA^2 \;=\;55^2 + \left(\frac{20}{\sqrt{3}}\right)^2 \;=\; 3025 + \frac{400}{3} \;=\;\frac{9475}{3}$

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