Questions: A 46.0-m guy wire is attached to the top of a 33.4-m antenna and to a point on the ground. How far is the point on the ground from the base of the antenna, and what angle does the guy wire make with the ground? The point on the ground is meters away. (Round to the nearest tenth as needed.)

Solution Steps

The problem involves a right triangle where the guy wire is the hypotenuse, the antenna is one leg, and the distance from the base of the antenna to the point on the ground is the other leg. We are given:

• The length of the guy wire: \( c = 46.0 \, \text{m} \)
• The height of the antenna: \( a = 33.4 \, \text{m} \)

To find the distance from the base of the antenna to the point on the ground, we use the Pythagorean theorem: \[ c^2 = a^2 + b^2 \] Solving for \( b \) (the distance on the ground): \[ b^2 = c^2 - a^2 \] \[ b = \sqrt{c^2 - a^2} \] Substituting the given values: \[ b = \sqrt{46.0^2 - 33.4^2} = \sqrt{2116 - 1115.56} = \sqrt{1000.44} \approx 31.63 \, \text{m} \]

To find the angle \( \theta \) that the guy wire makes with the ground, we use the cosine function: \[ \cos(\theta) = \frac{b}{c} \] \[ \theta = \cos^{-1}\left(\frac{b}{c}\right) \] Substituting the known values: \[ \theta = \cos^{-1}\left(\frac{31.63}{46.0}\right) \approx 0.8126 \, \text{radians} \] Converting radians to degrees: \[ \theta \approx 0.8126 \times \frac{180}{\pi} \approx 46.56^\circ \]

Final Answer