Question

a man standing in the field observes flying bird in his north at an angle of elevation of 30⁰ after 2.5 minutes he observes the bird in his south at an angle of elevation 60⁰. if the bird flies in a straight line all along a height of 60√3 meters find its height​

Answer 1

Answer:

let bird covers x mtr distance in 2 min

& distance from north point to the boy is y mtr

tan30= 50rt3/y

y=150

tan60= 50rt3/(x-150)

x=200 mtr

speed= 200/2*60 mtr/sec

= 200*18/120*5 km/hr

= 6 km/hr

Answer 2

Expert Answer:

Let O be the position of the bird, B be the position of the boy, G be the position of the girl and FG be the building at which the girl is standing.

BO= 100m , FG= 20m

In ∆OLB, = sin30° OL =50m

OM = OL-ML = OL-FG = 50-20 = 30m

In ∆OMG,

Thus, the distance of the bird from the girl is

42.3 m.

or

Let PQ be the tower and let PA and PB be its shadows when the altitudes of the sun are 45o and 30o respectively. Then, PAQ = 45o, PBQ = 30o, PQ = 50m, AB =x m.

In ∆APQ,

cot 45° = AP upon PQ

1=AP upon 50 = 50m

In ∆BPQ,

cot 30° = BP UPON PQ

ROOT3= x+50 upon 50

Thus, x= 50

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