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° and the angle of elevation of the top of section ‘A’ is 60°.

Based on the above information, answer the following questions:

1. Find the length of the wire from the point ‘O’ to the top of section ‘B’.    1
2. Find the length of the wire from the point ‘O’ to the top of section ‘A’.    1
3. 1. Find the distance AB.    2
      OR
   2. Find the area of Δ OРВ.    2

Answer 1

Given: OP = 6 m

∠ POB = 30°

∠ POA = 60°

(i) In Δ OРВ, a right-angled triangle.

∠ OPB = 90°

cos 30° = `(OP)/(OB)`

`sqrt3/2 = 6/(OB)`

`sqrt3 xx OB = 6 xx 2`

`OB = 12/sqrt3`

= `(12 xx sqrt3)/(sqrt3 xx sqrt3)`

= `(12 xx sqrt3)/3`

= `4sqrt3` m

(ii) In Δ APO, a right-angled triangle.

cos 60° = `(OP)/(OA)`

`1/2 = 6/(OA)`

OA = 6 × 2

= 12 m

(iii) (a) In Δ BPO, a right-angled triangle.

tan 30° = `(BP)/(OP)`

`\implies 1/sqrt(3) = (BP)/6`

`\implies` BP = `6/sqrt(3) xx sqrt(3)/sqrt(3)`

= `6/3 sqrt(3)`

= `2sqrt(3)` m

In Δ APO, a right-angled triangle. 

tan 60 = `(AP)/(OP)`

`sqrt3 = (AP)/6`

AP = `6sqrt3`

∴ AB = AP – BP

= `6sqrt3 - 2sqrt(3)` m

= `4sqrt(3)` m

= 4 × 1.73 m

= 6.92 m

OR

(b) In ΔBPO, a right-angled triangle.

tan 30° = `(BP)/(OP)`

`\implies 1/sqrt(3) = (BP)/36`

BP = `6/sqrt(3)`

= `6/sqrt(3) xx sqrt(3)/sqrt(3)`

= `(6 sqrt(3))/3`

= `2sqrt(3)` m

Now, Area of ΔOPB = `1/2 xx "height" xx "base"`

= `1/2 xx BP xx OP`

= `1/2 xx 2sqrt(3) xx 6`

= `6 sqrt(3)`

= 6 × 1.73

= 10.38 m²