Chapter 6, Problem 6P

Determining a Distance A surveyor has determined that a mountain is 2430 ft high. From the top of the mountain he measures the angles of depression to two landmarks at the base of the mountain and finds them to be 42º and 39°. (Observe that these are the same as the angles of elevation from the landmarks as shown in the figure at the left.) The angle between the lines of sight to the landmarks is 68º. Calculate the distance between the two landmarks.

To determine

The distance between the two landmarks.

Answer to Problem 6P

The distance between the two landmarks is $4194.3\text{ft}$

Explanation of Solution

Given:

The height of the mountain is $2430\text{ft}$ high. The angle of depression of the two landmarks is $42°$ and $39°$. The angle between the lines of sight to the landmarks is $68°$. The figure of mountain is given as,

Calculation:

Assume the triangle ABC as given below with the given specifications,

Figure (1)

Use trigonometric ratio of sine in triangle ABD to calculate the value of AB with respect to Figure (1),

$\begin{array}{c}\sin 42=\frac{2430}{AB}\left[\sin \theta =\frac{P}{H}\right]\\ AB=\frac{2430}{\sin 42}\\ =3631.58\end{array}$

The length of side AB is 3631.58.

Use trigonometric ratio of sine in triangle BDC to calculate the value of BC with respect to Figure (1),

$\begin{array}{c}\sin 39=\frac{2430}{BC}\left[\sin \theta =\frac{P}{H}\right]\\ BC=\frac{2430}{\sin 39}\\ =3861.31\end{array}$

The length of side BC is 3861.31.

The cosine law is,

$a^2=b^2+c^2-2bc\cos A$

Substitute AC for a, 3861.31 for b, and 3631.58 for c, and $68°$ for B to find the value of AC.

$\begin{array}{c}A{C}^2={\left(3861.31\right)}^2+{\left(3631.58\right)}^2-2\cdot \left(3861.31\right)\cdot \left(3631.58\right)\cdot \cos 68\\ =14909714.9616+13188373.2964-10505958.916\\ =17592129.339\end{array}$

Further solve the above expression.

$\begin{array}{c}A{C}^2=17592129.339\\ =\sqrt{17592129.339}\\ =4194.3\text{ft}\end{array}$

Thus, the distance between the two landmarks is $4194.3\text{ft}$.