Question

The diagram shows two vertical masts, AB and CD, on horizontal ground. The height of AB is 500m and the height of CD is 350m. The angle of elevation of B from D is 30 degrees. Calculate, to the nearest m, the distance AC.

Answer 1

Given:

The lines AB and CD, on horizontal ground.

The height of AB is 500 m.

The height of CD is 350 m.

The angle of elevation of B from D is 30°

We need to determine the distance AC.

Length of BE:

Let us construct a line that is parallel to AC.

Let the line be ED.

The length of BE is given by

$BE=AB-AE$

$BE=500-350$

$BE=150$

Thus, the length of BE is 150 m

Length of ED:

The length of ED can be determined using the trigonometric ratio.

Thus, we have;

$tan \ 30^{\circ}=\frac{ED}{BE}$

Substituting the values, we get;

         $\frac{\sqrt{3}}{3}=\frac{ED}{150}$

$\frac{\sqrt{3}}{3}\times 150=ED$

     $50\sqrt{3}=ED$

Thus, the length of ED is 50√3 m

Length of AC:

From the figure, it is obvious that the sides ED and AC have the equal length.

Thus, we have;

AC = ED = 50√3

Simplifying, we get;

$AC=50 \times 1.732$

$AC=86.6 \ m$

Hence, the length of AC is 86.6 m