From the top of a minar of height 60m,the top and bottom of a clock tower are observed at the angles of depression of 30 dgree and 60 degree respectively.Then the height of the clock tower in metres is A)40 B)50 C)60 D)80
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- Height of minar = 60m
- Depression angles are 30° & 60° respectively.
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Let AB is minar and EC is tower .
Given, AB = 60 m and EC = ?
In triangle ABC,
Let AB = x
⇒ BD = 60 – x
And, ∠ ACB = ∠OAC (alternate angles are equal)
∠ACB = 30°
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⇒We know that,
tan(theta) = p/b
tan 30° = AB/BC
1/√3=x/BC
BC=x√3............(i)
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⇒In ∆ADE,
∠ADE= ∠OAE(alternate angles are equal)
∠AED=60°
tan60° = AD/DE
√3=60/ DE
⇒DE = 60/√3.............(ii)
∵BC = DE
So, equation (i) = equation (ii)
x√3 = 60/√3
3x = 60
⇒x = 20m.
∵CE = BD and BD = 60 - x
CE = 60 -20
=40m
So, the height of clock tower is 40m.
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