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,

including microwave dishes. They are among the tallest human-made

structures. There are 2 main types: guyed and self-supporting

structures.

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 . Distance between

the base of the tower and point O is 36 m. From point O , the angle of

elevation of the top of section B is 30° and the angle of elevation of

the top of section A is 45°.
On the basis of the above information, answer any four of the

following questions:

(i) What is the height of the section B ?

(a) 12 √3 m (b) 12 √2 m (c) 8 √3 m (d) 4 √2 m

(ii) What is the height of the section A ?

(a) 12(2 - √2) (b) 24(2 - √2) (c) 12(3 - √3) (d) 24(3 - √3)

(iii) What is the length of the wire structure from the point O to the

top of section A ?

(a) 32 √2 m (b) 24 √3 m (c) 28 √3 m (d) 36 √2 m

(iv) What is the length of the wire structure from the point O to the

top of section B ?

(a) 12 √3 m (b) 24 √3 m (c) 28 √3 m (d) 16 √3 m

(v) What is the angle of depression from top of tower to point O ?

(a) 30° (b) 45° (c) 15° (d) 75°​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

Answer 2

Answer:

(a) height of section B = 20.78 m

(b) height of section A = 36 m

(c) Length of the wire structure from the point O to the top of section A  $36\sqrt{2} m$

d) Length of the wire structure from the point O to the top of section A  $12\sqrt{3} m$

e)  angle of depression from top of tower to point O is 75

Step-by-step explanation:

Refer to the diagram below

Given, that the distance between the base of the tower and point O = 36 m

Consider ΔOCB, tan 30 = $\frac{BC}{OC}$

$\frac{1}{\sqrt{3x} } =\frac{BC}{36}$

Hence, BC= 20.78 m

Let 'x' be the height AB

in ΔOAC, tan 45 = $\frac{AB+BC}{OC}$

Then, 36 = x+ 20.78

Implies, x = 36- 20.78

Hence, x = 15.22 m

Thus,  AC = 15.22+ 20.78 = 36 m

(a) height of section B (BO) = 20.78 m

(b) height of section A (AO) = 36 m

c) length of the wire structure from the point O to the top of the section $A$

$\Rightarrow \cos 45=36 /$Wire length to the top of the section $A$

$\Rightarrow 1 / \sqrt{2}=36 /$ Wire length to the top of the section $A$

$\Rightarrow$ wire length to the top of the section $A= 36 \sqrt{2}$

d) length of the wire structure from the point O to the top of the section $B$

$\Rightarrow \cos 30=36 /$Wire length to the top of the section $B$

$\Rightarrow \sqrt{3} / {2}=36 /$ Wire length to the top of the section $B$

$\Rightarrow$wire length to the top of the section $B= 12 \sqrt{3}$

e)  Angle of depression is cos 30+45=cos 75