The angle of elevation of a bird from a point 'h' metres above a lake is angle alpha and the angle of depression of its reflection in the lake is angle beta. Find the height of the bird.
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Answered by
14
Let a be a point h meters over the lake AF and B be the situation of the Bird.
Attract a line parallel to EF from A on BD at C.
But, BF = DF
Let, BC = m
so, BF = (m + h)
⇒ BF = DF = (m + h) metres
Consider ΔBAC,
AB = m cosec α ---------- (1)
and, AC = m cot α
Consider ΔACD,AC = (2h + m) cot β
Therefore, m cot α = (2h + m) cot β
⇒ m = 2h cot β / (cot α - cot β)
Substituting the value of m in (1) we get,
AB = cosec α [2h cot β / (cot α - cot β)] = 2h sec α / (tan β - tan α)
Now, 2h sec α / (tan β - tan α) = 2h AB/AC / (FB + h/AC - FB - h/AC)
= 2h*AB / (FB + h - FB + h)
= 2h*AB/2h
= AB is the Distance
Hope It Helps
And Fig Is Attached
Attract a line parallel to EF from A on BD at C.
But, BF = DF
Let, BC = m
so, BF = (m + h)
⇒ BF = DF = (m + h) metres
Consider ΔBAC,
AB = m cosec α ---------- (1)
and, AC = m cot α
Consider ΔACD,AC = (2h + m) cot β
Therefore, m cot α = (2h + m) cot β
⇒ m = 2h cot β / (cot α - cot β)
Substituting the value of m in (1) we get,
AB = cosec α [2h cot β / (cot α - cot β)] = 2h sec α / (tan β - tan α)
Now, 2h sec α / (tan β - tan α) = 2h AB/AC / (FB + h/AC - FB - h/AC)
= 2h*AB / (FB + h - FB + h)
= 2h*AB/2h
= AB is the Distance
Hope It Helps
And Fig Is Attached
Answered by
24
====================
HERE IS YOUR ANSWER ☞
====================
let's the hight = X m
__________-
GIVEN
=> EF = h meter
=> angle(AED) = alpha
and
=> angle(DEB) = beta
_____________
Now
=> AC = BC = X ____(low of reflection)
=> CD = EF = h
_____________
than
=> AD = AC - CD = (X - h)
and
=> BD = BC + CD = (X + h)
______________
In △DEB,
________________
In △AED
_________________
_________________
HENCE,
==========================
HOPE IT WILL HELP YOU ☺☺
==========================
DEVIL_KING ▄︻̷̿┻̿═━一
HERE IS YOUR ANSWER ☞
====================
let's the hight = X m
__________-
GIVEN
=> EF = h meter
=> angle(AED) = alpha
and
=> angle(DEB) = beta
_____________
Now
=> AC = BC = X ____(low of reflection)
=> CD = EF = h
_____________
than
=> AD = AC - CD = (X - h)
and
=> BD = BC + CD = (X + h)
______________
In △DEB,
________________
In △AED
_________________
_________________
HENCE,
==========================
HOPE IT WILL HELP YOU ☺☺
==========================
DEVIL_KING ▄︻̷̿┻̿═━一
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