Question

Find the length of the wire from the point 'O' to the top of section 'B'.

Answer 1

Step 1: Understanding the Concept:
In a right-angled triangle, the wire represents the hypotenuse connecting the ground point to the top of the section.
Step 2: Key Formula or Approach:
Use trigonometric ratio: \( \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \).
Step 3: Detailed Explanation:
Let \( P \) be the base of the tower. In right \(\Delta OPB\):
Base \( OP = 6 \text{ m} \).
Angle of elevation \( \angle BOP = 30^{\circ} \).
Let the length of the wire be \( OB \).
\[ \cos 30^{\circ} = \frac{OP}{OB} \]
\[ \frac{\sqrt{3}}{2} = \frac{6}{OB} \]
\[ OB = \frac{12}{\sqrt{3}} \]
Rationalizing:
\[ OB = \frac{12\sqrt{3}}{3} = 4\sqrt{3} \text{ m} \]
Step 4: Final Answer:
The length of the wire to the top of section 'B' is \(4\sqrt{3}\) m.

Answer 2

Step 1: Understanding the Concept:
Similarly, the wire to the top of section 'A' is the hypotenuse of the larger right triangle.
Step 2: Key Formula or Approach:
\[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \]
Step 3: Detailed Explanation:
In right \(\Delta OPA\):
Base \( OP = 6 \text{ m} \).
Angle of elevation \( \angle AOP = 60^{\circ} \).
Let the length of the wire be \( OA \).
\[ \cos 60^{\circ} = \frac{OP}{OA} \]
\[ \frac{1}{2} = \frac{6}{OA} \]
\[ OA = 12 \text{ m} \]
Step 4: Final Answer:
The length of the wire to the top of section 'A' is 12 m.

Answer 3

Step 1: Understanding the Concept:
The distance \( AB \) is the difference between the heights of the two sections.
Step 2: Key Formula or Approach:
\[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \]
\[ AB = AP - BP \]
Step 3: Detailed Explanation:
In \(\Delta OPB\):
\[ \tan 30^{\circ} = \frac{BP}{6} \implies BP = \frac{6}{\sqrt{3}} = 2\sqrt{3} \text{ m} \]
In \(\Delta OPA\):
\[ \tan 60^{\circ} = \frac{AP}{6} \implies AP = 6\sqrt{3} \text{ m} \]
Now, calculate \( AB \):
\[ AB = 6\sqrt{3} - 2\sqrt{3} \]
\[ AB = 4\sqrt{3} \text{ m} \]
Step 4: Final Answer:
The distance \( AB \) is \(4\sqrt{3}\) m.

Answer 4

Step 1: Understanding the Concept:
The area of a right-angled triangle is given by half the product of its base and perpendicular.
Step 2: Key Formula or Approach:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
Step 3: Detailed Explanation:
For \(\Delta OPB\):
Base \( OP = 6 \text{ m} \).
Height \( BP = 2\sqrt{3} \text{ m} \) (calculated in part iii-a).
\[ \text{Area} = \frac{1}{2} \times 6 \times 2\sqrt{3} \]
\[ \text{Area} = 6\sqrt{3} \text{ m}^2 \]
Step 4: Final Answer:
The area of \(\Delta OPB\) is \(6\sqrt{3}\) \text{ m}\^2.