From a point on the ground the elevation of a tower is \(60^\circ\). From another point \(10\,\text{m}\) vertically above the first, the elevation is \(45^\circ\). Find the height of the tower.
\(H=15+5\sqrt3\,\text{m}.\)
Step 1: Draw the situation
Let the height of the tower be \(H\) (in metres). Let the horizontal distance from the tower to the point on the ground be \(x\) (in metres).
Step 2: Use the first observation point (on the ground)
From the ground, the angle of elevation is \(60^\circ\). By definition of tangent,
\[ \tan 60^\circ = \dfrac{\text{opposite side}}{\text{adjacent side}} = \dfrac{H}{x} \]
We know \(\tan 60^\circ = \sqrt{3}\). So,
\[ H = \sqrt{3}x \]
Step 3: Use the second observation point (10 m above the ground)
The new point is 10 m higher. So, the vertical distance from this point to the top of the tower is \(H - 10\).
The angle of elevation is given as \(45^\circ\). Again using tangent,
\[ \tan 45^\circ = \dfrac{H - 10}{x} \]
But \(\tan 45^\circ = 1\). So,
\[ \dfrac{H - 10}{x} = 1 \quad \Rightarrow \quad x = H - 10 \]
Step 4: Substitute value of \(x\)
From Step 2: \(H = \sqrt{3}x\). Replace \(x\) by \(H - 10\):
\[ H = \sqrt{3}(H - 10) \]
Step 5: Solve the equation
\[ H = \sqrt{3}H - 10\sqrt{3} \]
Bring terms together:
\[ H - \sqrt{3}H = -10\sqrt{3} \]
\[ H(1 - \sqrt{3}) = -10\sqrt{3} \]
Divide both sides by \((1 - \sqrt{3})\):
\[ H = \dfrac{-10\sqrt{3}}{1 - \sqrt{3}} \]
Step 6: Simplify the fraction
Multiply numerator and denominator by \((1 + \sqrt{3})\) to remove the surd from the denominator:
\[ H = \dfrac{-10\sqrt{3}(1 + \sqrt{3})}{(1 - \sqrt{3})(1 + \sqrt{3})} \]
\[ H = \dfrac{-10\sqrt{3}(1 + \sqrt{3})}{1 - (\sqrt{3})^2} \]
\[ H = \dfrac{-10\sqrt{3}(1 + \sqrt{3})}{1 - 3} = \dfrac{-10\sqrt{3}(1 + \sqrt{3})}{-2} \]
\[ H = 5\sqrt{3}(1 + \sqrt{3}) \]
\[ H = 5\sqrt{3} + 15 \]
Final Answer: The height of the tower is \(H = 15 + 5\sqrt{3}\,\text{m}\).