A vertical tower 141 ft high is anchored to the ground by guy wires as shown in the illustration. How long is each guy wire? Round your answer to the nearest hundredth.
A) A = 170 ft, B = 130 ft, C = 92 ft, D = 73 ft, E = 110 ft, F = 150 ft.
B) A = 167.28 ft, B = 129.54 ft, C = 91.81 ft, D = 72.95 ft, E = 110.68 ft, F = 148.41 ft.
C) A = 167 ft, B = 130 ft, C = 92 ft, D = 73 ft, E = 111 ft, F = 148 ft.
D) A = 166 ft, B = 129 ft, C = 91 ft, D = 74 ft, E = 110 ft, F = 149 ft.
E) A = 170 ft, B = 128 ft, C = 92 ft, D = 73 ft, E = 108 ft, F = 150 ft.

Question

A vertical tower 141 ft high is anchored to the ground by guy wires as shown in the illustration. How long is each guy wire? Round your answer to the nearest hundredth. 
A) A = 170 ft, B = 130 ft, C = 92 ft, D = 73 ft, E = 110 ft, F = 150 ft. 
B) A = 167.28 ft, B = 129.54 ft, C = 91.81 ft, D = 72.95 ft, E = 110.68 ft, F = 148.41 ft. 
C) A = 167 ft, B = 130 ft, C = 92 ft, D = 73 ft, E = 111 ft, F = 148 ft. 
D) A = 166 ft, B = 129 ft, C = 91 ft, D = 74 ft, E = 110 ft, F = 149 ft. 
E) A = 170 ft, B = 128 ft, C = 92 ft, D = 73 ft, E = 108 ft, F = 150 ft.

Quizplus · Accepted Answer

Using trigonometry, we know that $	an	heta = \frac{AB}{BC}$ and $	an	heta = \frac{DE}{EC}$
Thus, $AB = BC	an	heta$ and $DE = EC	an	heta$
Using Pythagoras' theorem, we know that $BC^2 + AD^2 = 141^2$ and $EC^2 + AF^2 = 141^2$
Substituting $AB$ and $DE$ from above, we get $(BC	an	heta)^2 + AD^2 = 141^2$ and $(EC	an	heta)^2 + AF^2 = 141^2$
We also know that $BC = EC$ and $AD = AF$ (from the symmetry of the triangle)
So, $(BC	an	heta)^2 + (141^2 - BC^2) = (EC	an	heta)^2 + (141^2 - EC^2)$
Simplifying, we get $BC = EC = \frac{141}{\sqrt{1 + 	an^2	heta}}$
Substituting $	heta = \arctan\frac{AD}{BC}$, we get $BC = EC = \frac{141}{\sqrt{1 + (\frac{AD}{BC})^2}}$
Plugging in values, we get $BC = EC \approx 129.54$ (rounded to the nearest hundredth)

A Vertical Tower 141 Ft High Is Anchored to the Ground