Question

1. if a 35 - foot cable were run from the top of the pole and anchored to the ground at a distance from the pole, about how far away from the pole would it be anchored? 30ft 15ft

Explanation:

Step1: Apply Pythagorean theorem

In a right - triangle, \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse (length of the cable \(c = 35\) ft) and \(a\) is one leg (height of the pole \(a=30\) ft), and \(b\) is the other leg (distance from the pole, which we want to find).
So, \(b^{2}=c^{2}-a^{2}\)
Substitute \(c = 35\) and \(a = 30\) into the formula: \(b^{2}=35^{2}-30^{2}\)

Step2: Calculate \(35^{2}-30^{2}\)

We know that \(35^{2}=35\times35 = 1225\) and \(30^{2}=30\times30=900\)
Then \(b^{2}=1225 - 900=325\)

Step3: Find \(b\)

\(b=\sqrt{325}\)
Simplify \(\sqrt{325}=\sqrt{25\times13}=5\sqrt{13}\approx5\times3.606 = 18.03\approx18\) (rounded to the nearest whole number)

Answer:

The cable would be anchored about \(18\) feet away from the pole.