Question

The angle of elevation of top of a pole from two points A and B on the horizontal line lying on the
of the pole are observed to be 300 and 60°. If AB=100 m then height of the pole is​

Answer 1

Answer:

height of the pole is​ 50√3 m

Step-by-step explanation:

The angle of elevation of top of a pole from two points A and B on the horizontal line lying on the

of the pole are observed to be 30° and 60°. If AB=100 m then height of the pole is​

Let say C is Base of Pole & D is top of Pole

Let say BC = X  m  CD = Y

AC = AB + BC = 100 + X m

Tan 30° = CD/AC  =  Y/(100 + X)

=> 1/√3 =  Y/(100 + X)

=> Y = (100 + X)/√3

Tan 60° = CD/BC  =  Y/X

=> √3 =  Y/ X

=> Y = X√3

(100 + X)/√3 = X√3

=> 100 + X = 3X

=> 2X = 100

=> X  = 50

Y = 50√3

height of the pole is​ 50√3 m

Answer 2

Answer:

Height of the pole = 25$\sqrt{3}$ meters

Step-by-step explanation:

Let the pole be CD

Let the height of the pole = h meters.

Angle of elevation from A to the top C of the pole = 30°

Angle of elevation from B to the top C of the pole = 60°

The distance between A and B = 100 m

Refer to the figure at the attachment.

From Triangle ACD,

Tan 30° = h/AD

=> 1/$\sqrt{3}$ = h/AD

=> AD = h$\sqrt{3}$    

From Triangle BCD,

tan 60° = h/BD

$\sqrt{3}$ = h/BD

BD = h/$\sqrt{3}$

Given;

AB = 100 m

AD + DB = 100

h$\sqrt{3}$  + h/$\sqrt{3}$ = 100

(3h + h)/$\sqrt{3}$ = 100

4h = 100$\sqrt{3}$

h = 25$\sqrt{3}$ meters