A telephone pole is supported by two wires on opposite sides. At the top of the pole, the wires form an angle of 60°. On the ground, the ends of the wires are 15.0 m apart. One wire makes a 45° angle with the ground.
How long are the wires, and how tall is the pole? Express your answers to the nearest tenth of a metre.
Let's call the length of the wire that makes a 45° angle with the ground x, and the length of the other wire y.
Since the wires form a 60° angle at the top of the pole, we can use the law of cosines to relate the lengths of the wires:
y^2 = x^2 + 15^2 - 2*x*15*cos(60°)
y^2 = x^2 + 225 - 30x(cos(60°))
y^2 = x^2 + 225 - 15x
Since one wire makes a 45° angle with the ground, we can use trigonometry to find the lengths of the wires:
x = y*cos(45°)
x = 0.7071y
Substitute this into the equation we found earlier:
y^2 = (0.7071y)^2 + 225 - 15(0.7071y)
y^2 = 0.5y^2 + 225 - 10.6065y
0.5y^2 - 10.6065y + 225 = 0
Using the quadratic formula:
y ≈ 21.3 m
Substitute this back into the equation for x:
x = 0.7071(21.3)
x ≈ 15.1 m
Now, to find the height of the pole, we can use trigonometry again:
tan(60°) = height/x
height = x * tan(60°)
height ≈ 15.1 * tan(60°)
height ≈ 26.2 m
Therefore, the lengths of the wires are approximately 21.3 m and 15.1 m, and the height of the pole is approximately 26.2 m.