Question

FROM THE TOP OF THE BUILDING 15M HIGH,THEN THE ANGLE OF ELEVATION OF THE TOP OF A TOWER IS FOUND TO BE 30 DEG.FROM THE BOTTOM OF THE SAME BUILDING,THE ANGLE OF ELEVATION OF THE TOP OF TH TOWER IS FOUND TO BE 45 DEG.DETERMINE THE HEIGHT OF THE TOWER AND THE DISTANCE BETWEEN THE TOWER AND THE BUILDING [NO LINKS PLZ]

Answer 1

NICE QUESTION here is the answer.
CASE-1
Given
height of tower=15m
angle of elevation=30°
tan∅=P/B
let base of towers is x
tan∅=15/x
tan 30°=15/√3
1/√3=15/x
x=15/√3..........(1)
CASE 2
tan ∅=P/B
tan 45°=P/15√3
P=15/√3
Hence height of tower and sistance between both towers are found to be same 15√3 which is 26 m.

Answer 2

(INCOMPLETE ANSWER PLEASE WAIT)


hi there,

The answer might be a little at start but it's really easy, here you go.

ok I am sorry, but I posted answer before pasting the image,

but, you can draw a figure as follows,

let AB be the height of the the building and let EC be the height of the tower, therefore BC will be the distance between the building and the tower.

construction : draw a perpendicular from A to EC at point D

in quadrilateral ABCD
angle B = angle C = 90° ...(given)
AD is perpendicular to EC
therefore,
angle D = 90°
angle A = 90°
(remaining angle of a quadrilateral)

therefore quadrilateral ABCD is a rectangle.

now, let ED be x
therefore, in right angled ∆ EBC
angle B = 45°
angle C = 90°
therefore angle E = 45°

therefore, ∆ EBC is an isoceles ∆.