The angle of elevation of the top of a tower from a point is \(30^\circ\). If the observer moves \(20\,\text{m}\) towards the tower, the angle becomes \(45^\circ\). Find the height.
\(h=10(\sqrt{3}+1)\,\text{m}\).
Step 1: Assume variables.
Let the height of the tower be \(h\,\text{m}\).
Let the initial distance of the observer from the tower be \(x\,\text{m}\).
Step 2: Use the first situation (angle = \(30^\circ\)).
From the definition of tangent,
\(\tan\theta = \dfrac{\text{opposite side}}{\text{adjacent side}}\).
Here, opposite side = tower height = \(h\), adjacent side = distance = \(x\).
So, \(\tan30^\circ = \dfrac{h}{x}\).
We know \(\tan30^\circ = \dfrac{1}{\sqrt{3}}\).
Therefore, \(\dfrac{1}{\sqrt{3}} = \dfrac{h}{x} \Rightarrow h = \dfrac{x}{\sqrt{3}}.\)
Step 3: Use the second situation (angle = \(45^\circ\)).
Now the observer moves \(20\,\text{m}\) closer. So, new distance = \((x - 20)\,\text{m}\).
Again using tangent,
\(\tan45^\circ = \dfrac{h}{x - 20}\).
We know \(\tan45^\circ = 1\).
So, \(1 = \dfrac{h}{x - 20} \Rightarrow h = x - 20.\)
Step 4: Combine the two equations for \(h\).
From first situation: \(h = \dfrac{x}{\sqrt{3}}\).
From second situation: \(h = x - 20\).
Therefore, \(\dfrac{x}{\sqrt{3}} = x - 20.\)
Step 5: Solve for \(x\).
Multiply both sides by \(\sqrt{3}\):
\(x = (x - 20)\sqrt{3}\).
Expand: \(x = x\sqrt{3} - 20\sqrt{3}\).
Bring terms with \(x\) together: \(x\sqrt{3} - x = 20\sqrt{3}\).
Factor \(x\): \(x(\sqrt{3} - 1) = 20\sqrt{3}\).
So, \(x = \dfrac{20\sqrt{3}}{\sqrt{3} - 1}.\)
Simplify by multiplying numerator and denominator by \((\sqrt{3}+1)\):
\(x = 10(3 + \sqrt{3})\,\text{m}.\)
Step 6: Find height \(h\).
From \(h = x - 20\):
\(h = 10(3 + \sqrt{3}) - 20 = 10(\sqrt{3} + 1)\,\text{m}.\)
Final Answer: The height of the tower is \(10(\sqrt{3}+1)\,\text{m}\).