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Calculating height of a unknown point?
Posted by Steph on April 16, 2016 at 4:23 amHello everyone, I am trying to work out the height of a point which is unknown, lets call that point 2.
The information that I have is this:
Plus the information that pt.1 is 20m high.
Now my question is, how can I work out the height of point B using this information?
I know you have to use trig to work out side pt1pt3, but then what after that?
Thanks for the help.
rankin_file replied 8 years, 3 months ago 9 Members · 12 Replies 
12 Replies

Can everyone see the picture I posted up?
Its telling me: You need 3 posts to add links to your posts! This is used to prevent spam.does this apply to image links?
thanks

Howdy Steph, welcome to the site….
The problem you show is a typical ASA (angle side angle) equation to solve.

Some things are not clear about your problem:
I suspect the writer of the problem wants you to assume line 23 is level, or equivalently that 13 is plumb, but they should have said so.
Your posted text refers to Point B, which I can’t find in the diagram.
It is not clear to me what to do with the 1.1 and 1.2 meter numbers. If Pt 1 and Pt 2 really are at the instrument and target then those numbers are irrelevant. You may need one or both of those numbers if there is another point B below the target or instrument.
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Bill93, post: 367534, member: 87 wrote: wants you to assume line 23 is level, or equivalently that 13 is plumb
I meant to say:
wants you to assume line 23 is level AND 23 is plumb, thus making a right angle and defining the relationship between 2 and 3.
. 
Very sorry bill, I cannot work out how to edit my original question to change point B to pt2, when I wrote point B I meant pt.2. sorry!
Let me redraw the question again. This time it might actually make sense.

1. Pt 1 has an elevation of 30m, and your “instrument” is set up 1.2 m above that. So the height of instrument (or HI) at point 1 is 31.2 metres.
2. The method of calculating the unknown sides of a right angle triangle is called trigonometry. The vertical distance from the “instrument” at pt1 to the “target” at pt3 is given by the relationship (cosine75Ã¥Â¡)*350 = 90.6m. So the height of the target at pt 2 is 31.290.6= 59.4 metres.
3. The target at pt 2 is 1.1 metres above the mark, so the height of pt2 is 59.41.1=60.5 metres. That’s 60.5 metres below sea level.
PT1+1.2(cos75*350)1.1=PT2

Ok no one has asked the real question here …. how would you observe pt3 from pt1 in the first place? The BS is below the set up to start and then to turn a VA of 75*00’00”.
Just saying :totalstation: 
Maybe this is out of a high school trig book written by someone unfamiliar with surveying.

Mark is right, it should be 60.49m below sea level. The sketch is kind of confusing, though.

Holy Cow, post: 367553, member: 50 wrote: Maybe this is out of a high school trig book written by someone unfamiliar with surveying.
Perhaps you’re right, but the drawing points in a different direction for its source. Problems like this in trig books are usually stated in terms of “angle of elevation” or “angle of depression.” That would be either the angle at Pt 2 or the complement of the given angle at Pt 1. While an instrument height is sometimes given, it is very rare two see two of them in a single problem. Also, the detailed symbols in the sketch point to a surveying or engineering source.
Perhaps the original poster can provide her source.
Pedigogicly speaking, it’s not bad. Showing a point below sea level provokes thought and there is a hint of three dimensions.

Brian Lepore, post: 367545, member: 11478 wrote: Ok no one has asked the real question here …. how would you observe pt3 from pt1 in the first place? The BS is below the set up to start and then to turn a VA of 75*00’00”.
Just saying :totalstation:Too easy….. it’s an old Kern instrument and measures nadir angles instead of zenith angles…..and that folks, is your blast from the past for today….
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