Question

example 6
two points a and c, are on the same level ground as the foot of the pole, b. the angles of elevation of the top of the pole d from a and c are 30° and 48° respectively.
if the distance between a and c is 50 m, find bd, the height of the pole if:
i) a and c are on the opposite sides of the vertical pole,
ii) a and c are on the same side of the vertical pole

Explanation:

Step1: Let the height of the pole $BD = h$.

Let $AB = x$ and $CB=y$.
In right - triangle $ABD$, $\tan30^{\circ}=\frac{BD}{AB}$, so $x = h\cot30^{\circ}=\sqrt{3}h$. In right - triangle $BCD$, $\tan48^{\circ}=\frac{BD}{CB}$, so $y = h\cot48^{\circ}$.

Step2: Case i: When $A$ and $C$ are on opposite sides of the pole

We know that $x + y=50$. Substituting the values of $x$ and $y$, we get $\sqrt{3}h+h\cot48^{\circ}=50$. Since $\cot48^{\circ}=\frac{1}{\tan48^{\circ}}\approx\frac{1}{1.1106}\approx0.9$.
So $h(\sqrt{3}+ 0.9)=50$. Then $h=\frac{50}{\sqrt{3}+0.9}\approx\frac{50}{1.732 + 0.9}=\frac{50}{2.632}\approx19.0$.

Step3: Case ii: When $A$ and $C$ are on the same side of the pole

We know that $x - y = 50$. Substituting the values of $x$ and $y$, we get $\sqrt{3}h - h\cot48^{\circ}=50$.
$h(\sqrt{3}-\cot48^{\circ})=50$. Then $h=\frac{50}{\sqrt{3}-0.9}\approx\frac{50}{1.732 - 0.9}=\frac{50}{0.832}\approx60.1$.

Answer:

i) Approximately $19.0$ m
ii) Approximately $60.1$ m