Example 11 from Algebra and Trigonometry, 9.2 Sum and Difference Identities

Investigating a Guy-wire Problem

For a climbing wall, a guy-wire $R$ is attached 47 feet high on a vertical pole. Added support is provided by another guy-wire $S$ attached 40 feet above ground on the same pole. If the wires are attached to the ground 50 feet from the pole, find the angle $α$ between the wires. See Figure 8.

Two right triangles. Both share the same base, 50 feet. The first has a height of 40 ft and hypotenuse S. The second has height 47 ft and hypotenuse R. The height sides of the triangles are overlapping. There is a B degree angle between R and the base, and an a degree angle between the two hypotenuses within the B degree angle.

Figure 8

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
Let’s first summarize the information we can gather from the diagram. As only the sides adjacent to the right angle are known, we can use the tangent function. Notice that $\tan β = \frac{47}{50},$ and $\tan (β - α) = \frac{40}{50} = \frac{4}{5} .$ We can then use difference formula for tangent.
$\tan (β - α) = \frac{\tan β - \tan α}{1 + \tan β \tan α}$

Now, substituting the values we know into the formula, we have

$\begin{matrix} \frac{4}{5} & = & \frac{\frac{47}{50} - \tan α}{1 + \frac{47}{50} \tan α} \\ 4 (1 + \frac{47}{50} \tan α) & = & 5 (\frac{47}{50} - \tan α) \end{matrix}$
Use the distributive property, and then simplify the functions.

$\begin{matrix} 4 (1) + 4 (\frac{47}{50}) \tan α & = & 5 (\frac{47}{50}) - 5 \tan α \\ 4 + 3.76 \tan α & = & 4.7 - 5 \tan α \\ 5 \tan α + 3.76 \tan α & = & 0.7 \\ 8.76 \tan α & = & 0.7 \\ \tan α & \approx & 0.07991 \\ \tan^{- 1} (0.07991) & \approx & .079741 \end{matrix}$
Now we can calculate the angle in degrees.

$α \approx 0.079741 (\frac{180}{π}) \approx 4.57°$

How did it go?

Solution

Solution