Question

Q75 At 10.30 p.m. the angle of elevation of
the top of the pole from the end of its 12
cm long shadow was 45° and at 12.30
p.m. the length of shadow of the same
x p.m. to the top of the pole is 30°,
find the length of the shadow at 12.30
p.m.​

Answer 1

Answer:

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:

$12.00 - 12.30 = 30minutes$

And

$45 degree - 30degree = 15degree$

So the length is

$30 + 15 = 45$

Answer is 45