Question

Write a relation between \(d\) (the height of window) and \(y\).

Answer 1

Step 1: Understanding the Concept:
In the given diagram, the window is at point \(W\).
The distance \(AX\) represents the height of the window \(d\) from the ground (assuming \(A\) is the foot of the tank tower on the ground).
Triangle \(\triangle AWX\) is a right-angled triangle where \(AW = y\) is the hypotenuse and \(AX = d\) is the opposite side to the angle of depression of \(30^{\circ}\).
Step 2: Key Formula or Approach:
We use the trigonometric ratio of sine:
\[ \sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} \]
Step 3: Detailed Explanation:
In right-angled triangle \(\triangle AWX\):
The angle of depression is \(30^{\circ}\), so \(\angle AWX = 30^{\circ}\).
\[ \sin 30^{\circ} = \frac{AX}{AW} \]
Substituting the given values:
\[ \frac{1}{2} = \frac{d}{y} \]
Rearranging to find the relation:
\[ y = 2d \]
Step 4: Final Answer:
The relation between \(d\) and \(y\) is \(y = 2d\).

Answer 2

Step 1: Understanding the Concept:
The value \(h\) represents the height of the upper part of the water tank tower (\(BX\)) relative to the horizontal line of sight from the window \(W\).
Step 2: Key Formula or Approach:
We use the trigonometric ratio of tangent in \(\triangle BWX\):
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \]
Step 3: Detailed Explanation:
In right-angled triangle \(\triangle BWX\):
The angle of elevation is \(45^{\circ}\), and the horizontal distance \(XW\) is equal to the distance between the building and the tank, which is 54 m.
\[ \tan 45^{\circ} = \frac{BX}{XW} \]
Substituting the values:
\[ 1 = \frac{h}{54} \]
\[ h = 54 \text{ m} \]
Step 4: Final Answer:
The value of \(h\) is 54 m.

Answer 3

Step 1: Understanding the Concept:
The total height of the water tank is the sum of the portions above and below the horizontal level of the window, i.e., \(AB = BX + XA = h + d\).
Step 2: Key Formula or Approach:
We find \(d\) using the tangent ratio in \(\triangle AWX\):
\[ \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} \]
Step 3: Detailed Explanation:
From part (ii), we know \(h = 54\) m.
In right-angled triangle \(\triangle AWX\):
\[ \tan 30^{\circ} = \frac{AX}{XW} = \frac{d}{54} \]
\[ \frac{1}{\sqrt{3}} = \frac{d}{54} \]
\[ d = \frac{54}{\sqrt{3}} = \frac{54 \times \sqrt{3}}{3} = 18\sqrt{3} \text{ m} \]
Approximate value of \(d \approx 18 \times 1.732 = 31.176\) m.
Total height \(AB = h + d = 54 + 18\sqrt{3}\) m.
Total height \(\approx 54 + 31.18 = 85.18 \text{ m}\).
Step 4: Final Answer:
The height of the water tank is \(54 + 18\sqrt{3}\) m.

Answer 4

Step 1: Understanding the Concept:
The value \(x\) is the line-of-sight distance from the window to the top of the tank (\(BW\)).
The height of the window above ground level is the distance \(AX = d\).
Step 2: Detailed Explanation:
1. Finding \(x\):
In right-angled triangle \(\triangle BWX\):
\[ \cos 45^{\circ} = \frac{XW}{BW} = \frac{54}{x} \]
\[ \frac{1}{\sqrt{2}} = \frac{54}{x} \implies x = 54\sqrt{2} \text{ m} \]
2. Finding height of the window:
As calculated in part (iii)(a):
\[ \tan 30^{\circ} = \frac{d}{54} \implies d = 18\sqrt{3} \text{ m} \]
This \(d\) is the height of the window above ground level.
Step 3: Final Answer:
The value of \(x\) is \(54\sqrt{2}\) m and the height of the window is \(18\sqrt{3}\) m.