An observer \(1.5\) m tall is \(20.5\) m from a tower \(22\) m high. Find the angle of elevation of the top of the tower from the observer's eye.
\(\displaystyle \theta=\tan^{-1}\!\Big(\dfrac{22-1.5}{20.5}\Big)=\tan^{-1}\!\Big(\dfrac{20.5}{20.5}\Big)=\tan^{-1}(1)=45^\circ\).
Step 1: Write down the given information.
Step 2: The angle of elevation is measured from the observer's eye level, not from the ground. So, effective height of the tower (above the observer's eyes) is:
\(22 - 1.5 = 20.5\,\text{m}\).
Step 3: Now form a right triangle:
Step 4: Use the tangent formula:
\(\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{20.5}{20.5} = 1\)
Step 5: Find the angle from the trigonometric ratio:
\(\theta = \tan^{-1}(1) = 45^\circ\)
Final Answer: The angle of elevation of the top of the tower is \(45^\circ\).