From a point A, the angle of elevation of the top of a vertical tower situated on the roof of 50 m high building is found to be θ. After walking some distance towards the tower the angle of elevation of bottom of tower from point B is also found to be θ. Find the height of the tower
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Answered by
5
Hey, there!
Here's the answer you are looking for.
Please see the attachment (for reference).
As you can clearly see, x (height of tower), here, is (inversely) proportional to d (distance at which the person was standing).
So, there is no particular value of x, unless d is mentioned, because for every value of d, the value of theta will change and for every value of theta, there will be a different value of x.
Hope that helps.
Thanks.
Here's the answer you are looking for.
Please see the attachment (for reference).
As you can clearly see, x (height of tower), here, is (inversely) proportional to d (distance at which the person was standing).
So, there is no particular value of x, unless d is mentioned, because for every value of d, the value of theta will change and for every value of theta, there will be a different value of x.
Hope that helps.
Thanks.
Attachments:
Answered by
6
Here according to the question I have answered in the attachment.
The height of the tower will be=>
EC= (50a+50b)/b.
Remember some trigonmetric identity=>
1)Sin^2 + Cos^2 = 1
2)SinA =Perpendicular/hypotenuse
3)CosA= Base/Hypotenuse
4)TanA=SinA/CosA
or
Perpendicular/Base
5)SinA= 1/CosecA
6)CosA=1/SecA
7)TanA=1/CotA.
The height of the tower will be=>
EC= (50a+50b)/b.
Remember some trigonmetric identity=>
1)Sin^2 + Cos^2 = 1
2)SinA =Perpendicular/hypotenuse
3)CosA= Base/Hypotenuse
4)TanA=SinA/CosA
or
Perpendicular/Base
5)SinA= 1/CosecA
6)CosA=1/SecA
7)TanA=1/CotA.
Attachments:
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