Question

From the top of a building 15 m high the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower and the distance between the tower and building.

Answer 1

Answer:

The height of the tower is 22.5 m and the distance between tower and building is 12.9 m .

Step-by-step explanation:

Given :  

Height of the building, BC = AD = 15 m  

Let the distance between tower and building(AB) is  x m  & the Height of the tower be OA.  

Let DC = x m

Angle of elevation from the top of the  building to the top of the tower, ∠OCD = 30° & angle of elevation from the bottom of the same building , ∠OBA = 60°

In right triangle , ∆ODC,

tan 30° =  OD/DC

1/√3 = h/x

x = √3h ……….(1)

In right triangle , ∆OAB,

tan 60° = OA/AB

tan 60° = (AD + DO)/AB

√3 = (h + 15)/x

√3 = (h + 15)/(√3h)

[From eq 1]

√3 × √3h = h + 15  

3h = h + 15

3h - h = 15

2h = 15

h = 15/2

h = 7.5  

Distance between tower and building :  

x = √3h

x = 1.732 × 7.5  

x = 12.9  

Height of the tower ( OA) :  

OA = AD + OD

OA = 15 + h

OA = 15 + 7.5  

OA = 22.5 m  

Height of the tower = 22.5 m

Hence, the height of the tower is 22.5 m and the distance between tower and building is 12.9 m .

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

$\huge\bigstar\mathfrak\blue{\underline{\underline{SOLUTION:}}}$

Let AB be the building & CD be the tower.

Let the distance between tower & building be 'x'.

∠CAE= 30°&

∠CBD= 60°

AB= ED=15m

BD= AE= x

$tan60 \degree = \frac{CD}{BD} \\ \\ = > \sqrt{3} = \frac{CE+ ED}{x} \\ \\ = > x \sqrt{3} =CE + 15 \\ \\ = > CE= x \sqrt{3} - 15 \\ \\ \\ tan30 \degree = \frac{CE}{AE} = \frac{x \sqrt{3} - 15}{x} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{x \sqrt{3} - 15}{x} \\ \\ = > x = 3x - 15 \sqrt{ 3} \\ \\ = > - 2x = - 15 \sqrt{3} \\ \\ = > x = \frac{15 \sqrt{3} }{2} = \frac{15 \times 1.73}{2} \\ \\ = > x = \frac{25.95}{2} = 12.975m \\ \\ CE = x \sqrt{3} - 15 = \frac{15 \sqrt{3} }{2} \times \sqrt{3} - 15 \\ = > CE= \frac{15 \times 3}{2} - 15 = \frac{45}{2} - 15 \\ \\ = > CE = 7.5m$

CD= CE + ED= 7.5 +15= 22.5m

Height of tower CD= 22.5m distance between building & tower (x)= 12.975m.

hope it helps ☺️