The angle of elevation of the top of a tower is \(30^\circ\). If the height of the tower is doubled (observer fixed), then the angle of elevation also doubles.
False.
Step 1: Let the original height of the tower be \(h\) metres, and the horizontal distance of the observer from the base of the tower be \(d\) metres.
Step 2: By definition of tangent,
\[ \tan 30^\circ = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{h}{d} \]
So, \( \dfrac{h}{d} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}. \)
Step 3: Now, suppose the height of the tower is doubled. The new height = \(2h\) metres.
Let the new angle of elevation be \(\theta'\).
Again, using tangent:
\[ \tan \theta' = \dfrac{2h}{d} \]
Step 4: From Step 2, we know \( \dfrac{h}{d} = \tan 30^\circ. \)
So, \( \tan \theta' = 2 \times \tan 30^\circ. \)
\[ \tan \theta' = 2 \times \dfrac{1}{\sqrt{3}} = \dfrac{2}{\sqrt{3}} \approx 1.1547 \]
Step 5: Now find \(\theta'\):
\[ \theta' = \tan^{-1}(1.1547) \approx 49.4^\circ \]
Step 6: The new angle of elevation is about \(49.4^\circ\), not \(60^\circ\).
Therefore, the angle does not double when the height of the tower is doubled.