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Angle of elevation

Asked May 29, 2008, 10:31 PM — 5 Answers
If I know the angle of elevation of an object is 58 degrees at 50 feet from the object, is there any simple way of calculating what the angle of elevation would be 11 feet below the observation point at the same distance (50 feet) from the object ?

Any help would be appreciated.
Flashman2

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galactus's Avatar
galactus Posts: 2,272, Reputation: 1436
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#2

May 30, 2008, 05:42 AM
Yes, there is. You can use the law of tangents to find the height OP, then subtract 11.

OP=50tan(58)=80.02.

Now, you have PQ=80.02-11=69.02.

Now, you can use the law of tangents to find the angle you need.
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Flashman2's Avatar
Flashman2 Posts: 54, Reputation: 1
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#3

May 30, 2008, 09:19 PM
Thanks galactus, I'm not sure I asked the question properly. What I'm looking for is what the elevation would be 11 feet BELOW the observation point "A" in your drawing.
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Unknown008 Posts: 8,147, Reputation: 3745
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#4

May 31, 2008, 02:22 AM
Then it would be an angle of depression, right. But the same procedure and reasoning is used. Then however it's simpler as you already know the distance P to the 'new' Q in the above sketch, which is 11 feet and only have to use the law of tangents to find the angle of depression.
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Flashman2's Avatar
Flashman2 Posts: 54, Reputation: 1
Junior Member
  
#5

Jun 2, 2008, 05:53 PM
Thanks Galactus and Unknown 008. I have finally figured it out with your help. If I knew how to attach a drawing to my question ,like galactus did, it would have been a easier for me to show what I wanted. But I don't know as yet and would like help in this regard.
Thanks again.
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galactus's Avatar
galactus Posts: 2,272, Reputation: 1436
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#6

Jun 2, 2008, 06:42 PM
I make mine in Paint. Click on Manage Attachments, Browse, locate your file then Upload. That's all there is to it.
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