Two men on the same side of a tall building notice the angle of elevation to the top of the building to 30 and 60 respectively. If the height of the building is known to be h= 60m find the distance between two men
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A------------------------------B----------------------C
x y
let CD is the building and A & B are the two person. join the point A with D and B with D. from A elevation is 30 degree and from B elevation is 60 degree.
AB = distance between two person = x m
in triangle ACD
tan 30 = CD/AC
1/√3 = 60/AC
AC = 60√3
x+y = 60√3 -----------------(1)
in triangle BCD
tan 60 = CD/BC
√3 = 60/BC
BC = 60/√3
y = 60/√3
put the value of y in equation (1)
x + 60/√3 = 60√3
x = 60√3 - 60/√3
x = 60(3-1)/√3
x = 120/√3
x = 40√3 meter
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construct a right angled triangle ABC. right angle is at point B and top of triangle is A and point C is lie on base.consider a point D on base of triangle ABC.join point D with A,see line AD is inclined on line CB at 60 and line AC is inclined on CB at 30 degree.DB line is x meter and BC is y meter consider
In ΔADB
x/60=cot 60
x/60=1/√3
x=60/√3
In ΔABC
y/60=cot 30
y/60=√3
y=60√3
distance b/w men is y-x⇒60√3 -60/√3
180- 60 ⇒40√3 m
√3
In ΔADB
x/60=cot 60
x/60=1/√3
x=60/√3
In ΔABC
y/60=cot 30
y/60=√3
y=60√3
distance b/w men is y-x⇒60√3 -60/√3
180- 60 ⇒40√3 m
√3
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