Chapter 12.8, Problem 3P

To determine

Find the acute angle that the shorter cable makes with the ground and the distance between the cables along the ground.

Answer to Problem 3P

The acute angle that the shorter cable makes with the ground is about $55.7°$ and the distance between the cables along the ground is about 353 m.

Explanation of Solution

Given:

Two cables of length 300 m and 270 m extend from the top of a television antenna to the level ground on opposite sides of the antenna.The longer cable makes an angle of $48°$ with the ground.

Formula Used:

Law of Cosine :

  $a^2=b^2+c^2-2bc\cos A$

Law of Sine:

  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

  

Calculation:

Represent the situation :

  

We need to find measure of angle C and a

Use Law of Sine:

  $\begin{array}{l}\frac{270}{\sin 48°}=\frac{300}{\sin C}\\ \Rightarrow \sin C=\frac{\sin 48°}{270}\cdot \mathrm{300.................}[\text{multiply each side by }\sin C\cdot \frac{\sin 48°}{270}]\\ \Rightarrow C\approx 55.7°\\ \Rightarrow A=180°-48°-55.7°=76.3°\end{array}$

Use Law of Cosine:

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ \Rightarrow a^2={270}^2+{300}^2-2(270)(300)\cos 76.3°\\ \Rightarrow a^2=162900-38367.76\\ \Rightarrow a=\sqrt{124532.24}\\ \Rightarrow a\approx 352.89\approx 353\end{array}$

So,the acute angle that the shorter cable makes with the ground is about $55.7°$ and the distance between the cables along the ground is about 353 m.