Question

Radio station tower built in two sections ‘A’ and ‘B’. Tower is supported by wires from a point ‘O’. (Assuming standard height/angle data from such figures: Base OP = 36m, $\angle AOP = 30^\circ$, $\angle BOP = 45^\circ$).

(i) Length of wire from O to top of B.
(ii) Length of wire from O to top of A.
(iii) (a) Distance AB. OR
(iii) (b) Area of \(\triangle OPB\).}

Hint:

When $\theta = 45^\circ$ in a right triangle, the base and perpendicular are equal, and the hypotenuse is always $\text{side} \times \sqrt{2}$.

Answer 1

Step 1: Understanding the Concept:
This problem involves two right-angled triangles $\triangle OPA$ and $\triangle OPB$ with a common base $OP$. We use trigonometric ratios ($\cos$ for wire length, $\tan$ for height).
Step 2: Key Formula or Approach:
1. $\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} \implies \text{Wire} = \frac{\text{Base}}{\cos \theta}$. 2. $\tan \theta = \frac{\text{Opposite}}{\text{Base}} \implies \text{Height} = \text{Base} \times \tan \theta$.
Step 3: Detailed Explanation:
1. (i) Wire to B ($\theta = 45^\circ$): - $OB = \frac{36}{\cos 45^\circ} = \frac{36}{1/\sqrt{2}} = 36\sqrt{2}$ m. 2. (ii) Wire to A ($\theta = 30^\circ$): - $OA = \frac{36}{\cos 30^\circ} = \frac{36}{\sqrt{3}/2} = \frac{72}{\sqrt{3}} = 24\sqrt{3}$ m. 3. (iii) (a) Distance AB: - $PB = 36 \tan 45^\circ = 36(1) = 36$ m. - $PA = 36 \tan 30^\circ = 36(1/\sqrt{3}) = 12\sqrt{3}$ m. - $AB = PB - PA = 36 - 12\sqrt{3}$ m. 4. (iii) (b) OR Area $\triangle OPB$: - $\text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 36 \times 36 = 648$ m$^2$.
Step 4: Final Answer:
(i) $36\sqrt{2}$ m. (ii) $24\sqrt{3}$ m. (iii)(a) $36 - 12\sqrt{3}$ m or (iii)(b) 648 m$^2$.