Question 14, Exercise 8.1

Solutions of Question 14 of Exercise 8.1 of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

A telephone pole is braced by two wires that are both fastened to the ground at a point 3m from the base of the pole. The shorter wire is fastened to the pole 3m above the ground and the longer wire 7m above the ground.
a. What is the measure, in degrees, of the angle that the shorter wire makes with the ground?
b. Let $\theta$ be the measure of the angle that the longer wire makes with the ground. Find $\sin \theta$ and $\cos \theta$.
c. Find the cosine of the angle between the wires where they meet at the ground. d. Find, to the nearest degree, the measure of the angle between the wires.

Solution.

Let wires are fasten at point A on the ground. Pole is along BD.

a. If shorter wire makes an angle $\alpha$ with the ground as shown in figure, we have $$\begin{align*} &\tan\alpha = \frac{\overline{BC}}{\overline{AB}} \\ \implies &\tan\alpha = \frac{3}{3} = 1 \\ \implies &\alpha = \tan^{-1}(1) = 45^\circ \end{align*}$$ Hence wire will make an angle $45^\circ$ with the ground.

b. If $\theta$ is an angle that the longer wire makes with the ground, then by Pythagoras theorem $$\begin{align*} & \overline{AD}^2 = \overline{AB}^2+\overline{BD}^2 \\ \implies \overline{AD}^2 = 3^2 + 7^2 = 58 \\ \implies \overline{AD} = \sqrt{58} \\ \end{align*}$$ Now $$\begin{align*} \sin\theta = \frac{\overline{BD}}{\overline{AD}} = \frac{7}{\sqrt{58}} \end{align*}$$ $$\begin{align*} \cos\theta = \frac{\overline{AB}}{\overline{AD}} = \frac{3}{\sqrt{58}} \end{align*}$$

c. From the figure, we see $\theta-\alpha$ is and angle between the wires where they meet at the ground. Thus

$$\begin{align*} \cos(\theta - \alpha) &= \cos\theta \cos\alpha + \sin\theta \sin\alpha \\ & = \frac{3}{\sqrt{58}}\times \cos 45^\circ + \frac{7}{\sqrt{58}}\times \sin45^\circ \\ & = (0.39391)(0.70711) + (0.91915)(0.70711) \\ &=0.9285 \end{align*}$$

d. As $$\begin{align*} \cos(\theta - \alpha) & =0.9285 \\ \implies \theta - \alpha & = \cos^{-1}(0.9285)\\ & = 21.797 \approx 22^\circ \end{align*}$$ Hence the measure of the angle between the wires is $22^\circ$ approximately.

< Question 13