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: Recognise the special right-triangle ratio.
In right triangle $OPB$, the angle at $O$ is $30^{\circ}$, the right angle is at $P$, so the remaining angle at $B$ is $60^{\circ}$. This is the classic $30^{\circ}\text{-}60^{\circ}\text{-}90^{\circ}$ triangle, whose three sides are always in the fixed ratio $1 : \sqrt{3} : 2$, where the side opposite $30^{\circ}$ is the smallest, the side opposite $60^{\circ}$ is $\sqrt{3}$ times that, and the hypotenuse (opposite $90^{\circ}$) is twice the smallest side.

Step 2: Identify which given side corresponds to which ratio part.
The side $OP = 6\text{ m}$ is adjacent to the $30^{\circ}$ angle at $O$ and opposite the $60^{\circ}$ angle at $B$, so it corresponds to the "$\sqrt{3}$" part of the ratio.

Step 3: Find the scale factor $k$.
\[ k\sqrt{3} = 6 \implies k = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \]
Step 4: Use the scale factor to get the hypotenuse, which is the wire $OB$.
The hypotenuse corresponds to the "$2$" part of the ratio:
\[ OB = 2k = 2 \times 2\sqrt{3} = 4\sqrt{3}\text{ m} \]
Final Answer:
The length of the wire from point 'O' to the top of section 'B' is $4\sqrt{3}\text{ m}$. \[ \boxed{4\sqrt{3}\text{ m}} \]