Question

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

Answer 1

Step 1: Identify the Triangle and Required Side:
In right \( \triangle OPB \), the wire \( OB \) forms the hypotenuse.
We are given the base \( OP = 6 \text{ m} \) and the angle \( \angle BOP = 30^{\circ} \).

Step 2: Use the Cosine Ratio:
\[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \]
Step 3: Substitute the Known Values:
\[ \cos 30^{\circ} = \frac{6}{OB} \] \[ \frac{\sqrt{3}}{2} = \frac{6}{OB} \]
Step 4: Solve for \( OB \):
\[ OB = \frac{6 \times 2}{\sqrt{3}} \] \[ OB = \frac{12}{\sqrt{3}} \] Rationalizing the denominator: \[ OB = \frac{12\sqrt{3}}{3} \] \[ OB = 4\sqrt{3} \text{ m} \]
Step 5: Final Answer:
\[ \boxed{OB = 4\sqrt{3} \text{ m}} \]

Answer 2

Step 1: Recall the Required Trigonometric Ratio:
The wire \( OA \) forms the hypotenuse of right \( \triangle OPA \).
To find the hypotenuse when the base and angle are known, we use the cosine ratio.

Step 2: Apply the Cosine Formula:
\[ \cos \theta = \frac{\text{Base}}{\text{Hypotenuse}} \]
Step 3: Substitute the Given Values:
In \( \triangle OPA \):
Base \( OP = 6 \text{ m} \)
Angle \( \angle AOP = 60^{\circ} \)
Let \( OA \) be the hypotenuse.

\[ \cos 60^{\circ} = \frac{6}{OA} \] \[ \frac{1}{2} = \frac{6}{OA} \]
Step 4: Solve for \( OA \):
\[ OA = \frac{6}{1/2} \] \[ OA = 6 \times 2 \] \[ OA = 12 \text{ m} \]
Step 5: Final Answer:
\[ \boxed{OA = 12 \text{ m}} \]

Answer 3

Step 1: Identify the Required Relationship:
The vertical distance \( AB \) is the difference between the two heights \( AP \) and \( BP \).
So, we first calculate \( AP \) and \( BP \) using the tangent ratio.

Step 2: Use Tangent Formula:
\[ \tan \theta = \frac{\text{Perpendicular}}{\text{Base}} \]
Step 3: Find \( BP \) in \( \triangle OPB \):
\[ \tan 30^{\circ} = \frac{BP}{6} \] \[ \frac{1}{\sqrt{3}} = \frac{BP}{6} \] \[ BP = \frac{6}{\sqrt{3}} \] \[ BP = 2\sqrt{3} \text{ m} \]
Step 4: Find \( AP \) in \( \triangle OPA \):
\[ \tan 60^{\circ} = \frac{AP}{6} \] \[ \sqrt{3} = \frac{AP}{6} \] \[ AP = 6\sqrt{3} \text{ m} \]
Step 5: Calculate \( AB \):
\[ AB = AP - BP \] \[ AB = 6\sqrt{3} - 2\sqrt{3} \] \[ AB = 4\sqrt{3} \text{ m} \]
Step 6: Final Answer:
\[ \boxed{AB = 4\sqrt{3} \text{ m}} \]

Answer 4

Step 1: Identify the Required Formula:
Since \( \triangle OPB \) is a right-angled triangle, its area can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Perpendicular} \]
Step 2: Note the Given Measurements:
Base \( OP = 6 \text{ m} \)
Perpendicular \( BP = 2\sqrt{3} \text{ m} \)

Step 3: Substitute the Values:
\[ \text{Area} = \frac{1}{2} \times 6 \times 2\sqrt{3} \]
Step 4: Simplify the Expression:
\[ \text{Area} = 3 \times 2\sqrt{3} \] \[ \text{Area} = 6\sqrt{3} \]
Step 5: Final Answer:
The area of \( \triangle OPB \) is \[ \boxed{6\sqrt{3} \text{ m}^2} \]