Question

Radio towers are used for transmitting a range of communication services including radio and television. The tower will either act as an antenna itself or support one or more antennas on its structure. On a similar concept, a radio station tower was built in two sections 'A' and 'B'. Tower is supported by wires from a point 'O' (as shown in figure). Distance between the base of the tower and point 'O' is 6 m. From point 'O', the angle of elevation of the top of the section 'B' is $30^\circ$ and the angle of elevation of the top of section 'A' is $60^\circ$.

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

Hint:

Using standard rationalization rules like $\frac{12}{\sqrt{3}} = 4\sqrt{3}$ saves you from doing long divisions in exams.

Answer 1

Step 1: Understanding the Question:
Let the base of the tower be $P$.
The point $O$ is at a distance of $OP = 6\text{ m}$ from the base.
The angle of elevation to the top of section $B$ is $\angle POB = 30^\circ$.
We need to find the length of the wire $OB$.

Step 2: Key Formula or Approach:
In right-angled triangle $OPB$:
The side adjacent to the angle is the base $OP = 6\text{ m}$.
The hypotenuse is the length of the wire $OB$.
The trigonometric ratio relating the base and hypotenuse is cosine:
\[ \cos 30^\circ = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{OP}{OB} \]

Step 3: Detailed Explanation:

• 1. Set up the trigonometric equation:
\[ \cos 30^\circ = \frac{6}{OB} \]

• 2. Substitute the standard value $\cos 30^\circ = \frac{\sqrt{3}}{2}$:
\[ \frac{\sqrt{3}}{2} = \frac{6}{OB} \]

• 3. Solve for $OB$ using cross-multiplication:
\[ OB = \frac{12}{\sqrt{3}} \]

• 4. Rationalize the denominator:
\[ OB = \frac{12\sqrt{3}}{3} = 4\sqrt{3}\text{ m} \]


Step 4: Final Answer:
The length of the wire from point 'O' to the top of section 'B' is $4\sqrt{3}\text{ m}$.