How to deal with trigonometry in application?

Question

A utility pole is anchored to level ground by two 53 ft wires, one on each side of the pole. The wires make a 65 (degree) angle with the ground below. How far apart along the ground are the wires?

Answer 1

In this case, we know two angles and one side, so we can use the Law of Sines to solve the problem.

 

We know that angles A and B = 65^o, which means that angle C = 180^o - 65^o - 65^o = 50^o

 

the Law of Sines is that side a/sin angle A = side b/ sin angle B = side c/sine angle C

 

side A is 53ft, therefore, 53'/sin 65^o = x/sin 50^o

 

We get 53'/0.906 = x/0.766

rearranging, x = (0.766 x 53')/0.906

and the answer is 44.8'

Answer 2

the 2 wires and the ground between them form an isosceles triangle

the pole divides this triangle into 2 right triangles

take either right triangle

the pole is one leg, half the ground distance is the other leg, and the wire is the hypotenuse

cos=adjacent/hypotenuse

cos(65º)=(half the ground distance)/53

cos(65º)(53)=half the ground distance

(0.422618)(53)=half the ground distance (the distance from where the wire is anchored to the ground to the pole in the ground)

(0.422618)(53)=22.3987 feet from one wire on the ground to the pole

double this distance

2*22.3987=44.7974 feet rounded to the nearest tenth is 44.8 feet which is c)

Answer 3

This is a poorly-worded question.  I think the person devising the problem had in mind solving the total length of the shadows of both wires, if the sun were directly overhead (their perpendicular projection onto the ground plane). 

 

We have two right triangles, back-to-back.  Each hypotenuse is a wire, and the sum of the measurements of the lower legs is "how far apart along the ground" the wires are.  So, we'll solve for a lower leg, and then double its length to get the solution.

 

Imagine the leftmost triangle on the xy plane, with the wire and ground intersecting at the origin, and the point where the wire touches the telephone pole somewhere off to the upper right of the origin.  If we can find the x-coordinate of this point, we're half-way there...literally (as you'll soon see).

 

Let's review a bit of trig.  For a point that is distance r and an angle of θ from the origin, the x-coordinate is

 

r*cos(θ).  This is our lower leg. 

Here, r = 53, and

θ = 65 degrees.

 

But the question asks for twice this length, since we have two wires to consider, and a symmetrical situation.

 

Our solution is

2r*cos(θ)

=2(53)cos(65 degrees)

=44.798 (to 3 decimal places)

 

Round and pick your answer.

 

Answer 4