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hope this helps you☺☺☺☺
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anugyasoni:
cot theta kaha se aya
tan 90- theta hota h cot theta
ohk ohk i got it thanks a lot
wlcm
thanks for mark☺☺
wr welcome
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Hey here is the solution
Let AB be the tower . Let C and D be the two points at distances a & b respectively from the base of the tower . Then ,AC=a and AD = b . Let angle ACB = θ and angle ADB = 90-θ .
Let h be the height of the tower AB.
In triangle CAB, We have
tanθ = AB/AC
tanθ = h/a ------->(1)
In triangle DAB , We have
tan(90-θ)=AB/AD
cotθ=h/b -------->(2)
From (1) & (2) we get
tanθ × cotθ = h²/ab
1= h²/ab
h²=ab
h=√ab metres
Hence Proved .
For diagram see the attachment
Let AB be the tower . Let C and D be the two points at distances a & b respectively from the base of the tower . Then ,AC=a and AD = b . Let angle ACB = θ and angle ADB = 90-θ .
Let h be the height of the tower AB.
In triangle CAB, We have
tanθ = AB/AC
tanθ = h/a ------->(1)
In triangle DAB , We have
tan(90-θ)=AB/AD
cotθ=h/b -------->(2)
From (1) & (2) we get
tanθ × cotθ = h²/ab
1= h²/ab
h²=ab
h=√ab metres
Hence Proved .
For diagram see the attachment
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