====== 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. 

===== Question 14=====
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. **

{{ :math-11-nbf:sol:unit08:math-11-nbf-ex8-1-q14.svg |Question 14: 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. GOOD
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