From two points A and B on the same side of a building, the angles of elevation of the top of the building are 30° and 60° respectively. If the height of the building is 10 m, find the distance between A and B, correct to two decimal places.

Let the height of the building be CD = 10 m,

Let the distance from the base of the building to point B be x and point A be y.

In right angled triangle CBD,

$\Rightarrow \tan \theta = \dfrac{\text{perpendicular}}{\text{base}} \\[1em] \Rightarrow \tan 60° = \dfrac{CD}{BD} \\[1em] \Rightarrow \sqrt{3} = \dfrac{10}{x} \\[1em] \Rightarrow x = \dfrac{10}{\sqrt3}.$

In right angled triangle CBD,

$\Rightarrow \tan 30° = \dfrac{CD}{AD} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{10}{y} \\[1em] \Rightarrow y = 10\sqrt{3}.$

The distance between the two points is the difference between their distances from the tower (y - x):

$\Rightarrow AB = y - x \\[1em] \Rightarrow AB = 10\sqrt3 - \dfrac{10}{\sqrt3} \\[1em] \Rightarrow AB = \dfrac{30 - 10}{\sqrt3} \\[1em] \Rightarrow AB = \dfrac{20}{\sqrt3} \\[1em] \Rightarrow AB = \dfrac{20}{1.732} \\[1em] \Rightarrow AB = 11.547 \approx 11.55 \text{ m}.$

Hence, the distance between A and B is 11.55 m.