Civil Surveying Sample Exams, Exam 1, Question 30

I'm having some trouble understanding the solution to this problem completely! Can anyone kindly walk me through it?

The problem reads

For the region shown, The following parameters apply.

(y,x) coordinate of A (1675.24, 2546.09)

(y,x) coordinate of B (1294.39, 2870.06)

azimuth of AP 95°34'20"

azimuth of BP 48°00'00"

palv,

There is more of the problem you did not include (like the actual question and the figure given). The problem is asking for the y,x coordinates at point P. You are given the coordinates of Pt A and Pt B and the azimuth of AP and BP. So, you can easily calc Angle P. You can then figure out the Angles at A and B pretty easily with trig (SohCahToa ). And you can easily figure out the length of AB. Now you know 3 angles and one side. This gives you enough info to use the Law of Sines to find more info. Does that help?

ptatohed said:

palv,

There is more of the problem you did not include (like the actual question and the figure given). The problem is asking for the y,x coordinates at point P. You are given the coordinates of Pt A and Pt B and the azimuth of AP and BP. So, you can easily calc Angle P. You can then figure out the Angles at A and B pretty easily with trig (SohCahToa ). And you can easily figure out the length of AB. Now you know 3 angles and one side. This gives you enough info to use the Law of Sines to find more info. Does that help?

Click to expand...

Ok thanks. Looks like I did miss writing the actual question.

I get how to solve for length AB. I now see how to solve for angle P. However, I'm not sure how do I solve for the angles at A and B using SohCahToa? Where is the right angle that would make SohCahToa applicable?

The right triangle is formed with Line AB being the hypotenuse. The ‘base’ is 2870.06 – 2546.09 and the ‘height’ is 1675.24 – 1294.39. The length of AB is, of course, (x2 + y2)½. This right triangle is actually outside of the enclosed traverse, but that’s ok. Once you get the exterior angle(s), you can easily get the interior angle(s) at A and B. Does that make sense?

What they want you to do here is find the coordinate value at point P by using what is known as a Bearing-Bearing intersection. First, inverse between A and B to get your bearing and distance between those two then Use the law of cosines to find the distances from A to P and B to P and then coordinate point P with the given azimuths. The solution will become apparent if you plot the coordinates up on a sheet of graph paper and then use a protractor to draw the lines represented by the given azimuths. That is why its called a bearing-bearing intersection. There is no right triangle involved.

canvasbak66 said:

What they want you to do here is find the coordinate value at point P by using what is known as a Bearing-Bearing intersection. First, inverse between A and B to get your bearing and distance between those two then Use the law of cosines to find the distances from A to P and B to P and then coordinate point P with the given azimuths. The solution will become apparent if you plot the coordinates up on a sheet of graph paper and then use a protractor to draw the lines represented by the given azimuths. That is why its called a bearing-bearing intersection. There is no right triangle involved.

Click to expand...

Hmm, I used Law of Sines. I did not think to use Law of Cosines. To me, you have more knowns for Sines than Cosines but either would work I guess. How does the book's solution do it? And, yes there is a right triangle involved. Show me your solution and I'll show you where you used a right triangle.

Got it. I see Ptatohed's way. I can also follow canvasbak66's solution (except I use law of sines). Thank you both.

I also found another way, though not as convenient for the exam because it is time consuming .... It involves setting up 4 equations with 4 unknowns as follows.

AX + LAP*Sin AZAP = PX

AY + LAP*Cos AZAP = PY

BX+ LBP*Sin AZBP = PX

AX + LBP*Cos AZBP = PY

AX, AY, BX, and BY are given as are AZAP and AZBP.... This method is more 8th grade algebra than anything...

If one assumes AP = L1 and PB = L2

Azimuth AP = 95.572 deg

Azimuth BP = 48.0 deg

Azimuth PB = back azimuth of BP = 180 + 48 = 228 deg

Going from A to B via AP and then PB:

change in northing = L1 cos 95.572 + L2 cos 228 = 1294.39 – 1675.24 = - 380.85

change in easting = L1 sin 95.572 + L2 sin 228 = 2870.06 – 2546.09 = + 323.97

-0.097L1 -0.669L2 = - 380.85

0.995 L1 – 0.743L2 = 323.97

Solving: L1 = 678.19ft; L2 = 471.1 ft

Going from A to P via AP:

change in northing = L1 cos 95.572 = 678.19 x cos 95.572 = -65.85

change in easting = L1 sin 95.572 =678.19 x sin 95.572 = + 674.99

Coordinates of P (y,x) = 1675.24 – 65.85 = 1609.39 , 2546.09 + 674.99 = 3221.08

Pretty Long!

Yeah, you are right c_n, it is a long problem. Certainly not a 2.72 minute problem.

I looked at the book's solution tonight and I can't really follow it. They state "The equations are...", followed by equations I have never seen before for yP and xP. ?

This is how I did it (3 years ago):

Solve for Angle P: 95o34'20" - 48o = 47o34'20" = 47.5722o

Solve for length AB: [(2870.06 - 2546.09)2 + (1675.24 - 1294.39)2].5 = 500.00'

Solve for Angle B: 48o + [90o - tan-1((1675.24 - 1294.39) / (2870.06 - 2546.09))] = 88.3861o

Solve for Angle A: 180o - (47.5722o + 88.3861o) = 44.0417o

Use Law of Sines to calc length BP: 500 = [bP / (sin (44.0417o))] [sin (47.5722o)]; BP = 470.91'

Calc the x and y distances from B to P: yBP = sin (42o) 470.91' = 315.10 ; xBP = cos (42o) 470.91' = 349.95

Add yBP and xBP to the coordinates at B: 2870.06 + 349.95 = 3220.01(x) and 1294.39 + 315.10 = 1609.49

Answer = (1609.49 , 3220.01) (y,x) = Answer D.

The law of sines is the correct formula to use.

Point P:

Northing = 1609.49

Easting = 3220.01