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 \(BD = h\) meters.

Let \(AB=x\) and \(CB = y\).

Step2: Use tangent function in right - triangles

In right - triangle \(ABD\), \(\tan30^{\circ}=\frac{BD}{AB}\), so \(\frac{1}{\sqrt{3}}=\frac{h}{x}\), then \(x = h\sqrt{3}\). In right - triangle \(CBD\), \(\tan48^{\circ}=\frac{BD}{CB}\), so \(y=\frac{h}{\tan48^{\circ}}\).

Case (i): When \(A\) and \(C\) are on the opposite sides of the vertical pole

Step3: Set up the equation based on the distance between \(A\) and \(C\)

We know that \(x + y=50\), substituting \(x = h\sqrt{3}\) and \(y=\frac{h}{\tan48^{\circ}}\) into the equation, we get \(h\sqrt{3}+\frac{h}{\tan48^{\circ}}=50\).
Factor out \(h\): \(h(\sqrt{3}+\frac{1}{\tan48^{\circ}})=50\).
Since \(\tan48^{\circ}\approx1.1106\), then \(\sqrt{3}+\frac{1}{\tan48^{\circ}}\approx1.732 + \frac{1}{1.1106}\approx1.732+0.9004 = 2.6324\).
So \(h=\frac{50}{2.6324}\approx19.0\) m.

Case (ii): When \(A\) and \(C\) are on the same side of the vertical pole

Step4: Set up the equation based on the distance between \(A\) and \(C\)

We know that \(x - y = 50\), substituting \(x = h\sqrt{3}\) and \(y=\frac{h}{\tan48^{\circ}}\) into the equation, we get \(h\sqrt{3}-\frac{h}{\tan48^{\circ}}=50\).
Factor out \(h\): \(h(\sqrt{3}-\frac{1}{\tan48^{\circ}})=50\).
Since \(\sqrt{3}-\frac{1}{\tan48^{\circ}}\approx1.732-0.9004 = 0.8316\).
So \(h=\frac{50}{0.8316}\approx60.1\) m.

Answer:

i) Approximately \(19.0\) m
ii) Approximately \(60.1\) m