Of course we can't show figures on the website, but I am assuming
this question is of the standard type.
Assume that the guy wires are attached at the same height up the tower.
Call the points where the wires are anchored on the ground as A and B
so we know AB = 150 ft.
Let the base of the tower be C and its top as D
Using standard geometry, angle ABD = 180-58 or 122°,
then angle ADB = 9°
The shorter wire would be DB
By the sine law:
150/sin9° = DB/sin49°
DB = 150sin49/sin9 = 723.67 ft
Normally this question continues to find the height of the tower or more
interesting facts
Two guy wires for the radio tower of SPI station made 58° and 49° angles with the horizontal as shown in the figure. If the ground anchors for each wire were 150 ft apart and assuming that the wires are straight, find the length of the shorter wire
2 answers
No figure shown, so there's something left undetermined.
Suppose we label things as follows:
T = top of pole
P = bottom of pole
so the pole's height h = PT
A = wire 1 anchor point (58°)
B = wire 2 anchor point (49°)
x = PA
y = PB
z = AB = 150
Then, if ∡APB is θ, we have
x^2 + y^2 - 2xy cosθ = 150
h/x = tan58°
h/y = tan49°
Since we don't know θ, we can't solve this yet. So, assuming the wires are anchored on opposite sides of the pole, we have
h cot58° + h cot49° = 150
h = 100.39
so the length q of the shorter wire is
h/q = sin58°
q = 118.38
Suppose we label things as follows:
T = top of pole
P = bottom of pole
so the pole's height h = PT
A = wire 1 anchor point (58°)
B = wire 2 anchor point (49°)
x = PA
y = PB
z = AB = 150
Then, if ∡APB is θ, we have
x^2 + y^2 - 2xy cosθ = 150
h/x = tan58°
h/y = tan49°
Since we don't know θ, we can't solve this yet. So, assuming the wires are anchored on opposite sides of the pole, we have
h cot58° + h cot49° = 150
h = 100.39
so the length q of the shorter wire is
h/q = sin58°
q = 118.38