The angles of elevation of the top of a tower from two points \(s\) and \(t\) metres from its foot are complementary. Prove that the height is \(\sqrt{st}\).
\(h=\sqrt{st}\).
Step 1: Understand the situation.
We have a vertical tower of height \(h\,\text{metres}\). Two people stand on the ground, at distances \(s\,\text{metres}\) and \(t\,\text{metres}\) away from the foot of the tower. They look up to the top of the tower, and their angles of elevation are complementary. This means that if one angle is \(\alpha\), the other is \(90^\circ - \alpha\).
Step 2: Write trigonometric relations.
For the person at distance \(s\):
\[ \tan(\alpha) = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{h}{s}. \]
For the person at distance \(t\):
The angle here is \(90^\circ - \alpha\). But we know that: \[ \tan(90^\circ - \alpha) = \cot(\alpha). \]
So, \[ \tan(90^\circ - \alpha) = \dfrac{h}{t} = \cot(\alpha). \]
Step 3: Relating the two equations.
From the first equation: \[ \tan(\alpha) = \dfrac{h}{s}. \]
From the second equation: \[ \cot(\alpha) = \dfrac{h}{t}. \]
We know that: \[ \tan(\alpha) \times \cot(\alpha) = 1. \]
Substitute the values: \[ \left( \dfrac{h}{s} \right) \times \left( \dfrac{h}{t} \right) = 1. \]
Step 4: Simplify the equation.
\[ \dfrac{h^2}{st} = 1. \]
Multiply both sides by \(st\):
\[ h^2 = st. \]
Step 5: Take square root.
\[ h = \sqrt{st}. \]
So, the height of the tower is \(\sqrt{st}\,\text{metres}.\)