Question

3. A telephone pole is supported by two wires on opposite sides. At the top of the pole, the wires
form an angle of 55º. On the ground, the ends of the wires are 11.0 m apart. One wire makes a
(4) 41° angle with the ground.
a) How long is each wire? (round to the nearest tenth)
b) How tall is the pole? (round to the nearest tenth)

I understand A, but please help with B ASAP!!!!

Answer 1

Answer: a) length of the wires are 8.1m and 13.4m

b) length of pole is 8.8m

Step-by-step explanation:

The diagram representing the scenario is shown in the attached photo. Triangle ABC is formed. Since the sum of the angles in a triangle is 180°, it means that angle B would be

180 - (41 + 55) = 84°

a) AC and BC are the lengths of each wire.

CD is the height of the pole

To determine AC, we would apply sine rule which is expressed as

a/SinA = b/SinB = c/SinC

Therefore,

11/Sin55 = BC/Sin 41 = AC/Sin84

1) 11//Sin 55 = BC/Sin 41

Cross multiplying, it becomes

11Sin41 = BCSin55

11 × 0.6561 = BC × 0.8192

BC = 7.2171/0.8192 = 8.8m

2) 11/Sin55 = BC/Sin 41 = AC/Sin84

1) 11//Sin 55 = AC/Sin 84

Cross multiplying, it becomes

11Sin84 = ACSin55

11 × 0.9945 = AC × 0.8192

AC = 10.9395/0.8192

AC = 13.4m

b) triangles ACD and BCD are right angle triangles. To determine CD, we would apply the sine trigonometric ratio.

Sin # = opposite side/adjacent side

Considering triangle ACD,

Sin 41 = CD/13.4

CD = 13.4Sin41 = 13.35 × 0.6561

CD = 8.8m