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: Reframing the Geometry:
Instead of directly applying individual trigonometric ratios, we carefully analyze the structure of the two right-angled triangles ΔOPA and ΔOPB.

• OP = 36 m (common horizontal base)
• Angle of elevation to B = 45°
• Angle of elevation to A = 30°
• OB and OA are hypotenuse (wire lengths)

Both triangles share the same base but differ in angle, so we use exact trigonometric values of special angles.

Step 2: Standard Trigonometric Values Used:
sin 45° = cos 45° = 1/√2
tan 45° = 1

sin 30° = 1/2
cos 30° = √3/2
tan 30° = 1/√3

Step 3: Detailed Calculations:

(i) Length of wire OB (θ = 45°)
Using:
cos θ = Base / Hypotenuse
⇒ Hypotenuse = Base / cos θ

OB = 36 / cos 45°
= 36 / (1/√2)
= 36√2 m

(ii) Length of wire OA (θ = 30°)
OA = 36 / cos 30°
= 36 / (√3/2)
= 72 / √3

Rationalising:
= (72√3) / 3
= 24√3 m

(iii)(a) Distance AB (difference in heights)
Height at B:
PB = 36 tan 45°
= 36 × 1
= 36 m

Height at A:
PA = 36 tan 30°
= 36 × (1/√3)
= 12√3 m

Therefore,
AB = PB − PA
= 36 − 12√3 m

(iii)(b) Area of triangle OPB
Area = 1/2 × Base × Height
= 1/2 × 36 × 36
= 648 m²

Final Answers:
(i) 36√2 m
(ii) 24√3 m
(iii)(a) 36 − 12√3 m
(iii)(b) 648 m²