Question

A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenna and anchored to the ground, at two anchor points on opposite sides of the antenna. The shorter wire makes an angle of 63º with the ground. How far apart are the anchor points?

Answer 1

Answer:

163.8 ft

Explanation:

In triangle ABD

$AB$ = 155 ft

$Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft$

$Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft$

Using Pythagorean theorem in triangle ADC

$AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft$

$d$ = distance between the anchor points

distance between the anchor points is given as

$d = BD + CD = 70.4 + 93.4\\d = 163.8 ft$

Answer 2

Answer:

far apart are the anchor points= 107.4866 ft

Explanation:

To solve this question we need to draw two right triangles, back to back. And the antenna is shared side in both triangle.  

Now, The triangle made by the shorter wire has a hypotenuse 155 feet that is the length of shorter side.

Similarly,  the triangle made by the longer wire has a hypotenuse 175 feet long (length of the longer wire).

The vertex at the shorter wire's anchor point is 63°.

Let us find the length of the leg from the shorter wire's anchor point to the base of the antenna.

We will use cos ratio

cos =base/hypotenuse

$cos63=\frac{x}{155}$

x=70.36 ft

Hence, the distance from the shorter wire's anchor point to the base of the antenna is 70.36 feet.

to find the height of antenna using longer wire triangle

sin63°= $\frac{h}{155}$

h=138.10 ft

Now, we can use the Pythagorean Theorem to find the third side of the longer wire's triangle, let that be a

$175^2= 138.10^2+a^2$

solving for a we get

a= 107.4866 ft

so anchor points are a= 107.4866 ft