Find: ZenithDA SDDA = 155.3 [ft] 89 29’15” 89 46’30” D 90 14’30” A BS HI FS Elev A 3.27 ElevA HIAB dd B 4.56 4.31 ElevB HIBC dd C 5.34 4.28 31.69 HICD dd D 5.97 ElevD SDDA = 155.3 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” o Find the zenith angle D A. [pause] In this problem, a level survey was performed --- D o A plan view o o B C

Find: ZenithDA SDDA = 155.3 [ft] 89 29’15” 89 46’30” D 90 14’30” A BS HI FS Elev A 3.27 ElevA HIAB dd B 4.56 4.31 ElevB HIBC dd C 5.34 4.28 31.69 HICD dd D 5.97 ElevD SDDA = 155.3 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” o between Points A, B, C and D. The backsight readings and forsight readings --- D o A plan view o o B C

Find: ZenithDA SDDA = 155.3 [ft] 89 29’15” 89 46’30” D 90 14’30” A BS HI FS Elev A 3.27 ElevA HIAB dd B 4.56 4.31 ElevB HIBC dd C 5.34 4.28 31.69 HICD dd D 5.97 ElevD SDDA = 155.3 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” o have all been recorded, but the only elevation provided is at Point C. --- D o A plan view o o B C

Find: ZenithDA SDDA = 155.3 [ft] 89 29’15” 89 46’30” D 90 14’30” A BS HI FS Elev A 3.27 ElevA HIAB dd B 4.56 4.31 ElevB HIBC dd C 5.34 4.28 31.69 HICD dd D 5.97 ElevD SDDA = 155.3 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” o [pause] The problem statement provides the slope distance between --- D o A plan view o o B C

Find: ZenithDA SDDA = 155.3 [ft] 89 29’15” 89 46’30” D 90 14’30” A BS HI FS Elev A 3.27 ElevA HIAB dd B 4.56 4.31 ElevB HIBC dd C 5.34 4.28 31.69 HICD dd D 5.97 ElevD SDDA = 155.3 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” o Points D and A, but no height of instrument values are provided. --- D o A plan view o o B C

Find: ZenithDA SDDA = 155.3 [ft] 89 29’15” 89 46’30” D 90 14’30” A BS HI FS Elev A 3.27 ElevA HIAB dd B 4.56 4.31 ElevB HIBC dd C 5.34 4.28 31.69 HICD dd D 5.97 ElevD SDDA = 155.3 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” o so we’ll remove this column from our data table, at least for the time being. ---- D o A plan view o o B C

Find: ZenithDA SDDA = 155.3 [ft] 89 29’15” 89 46’30” D 90 14’30” A BS FS Elev A 3.27 ElevA B 4.56 4.31 ElevB C 5.34 4.28 31.69 D 5.97 ElevD SDDA = 155.3 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” o [pause] Also, all length measurements recorded in the data table --- D o A plan view o o B C

Find: ZenithDA all lengths are in feet SDDA = 155.3 [ft] 89 29’15” BS FS Elev are in feet A 3.27 ElevA B 4.56 4.31 ElevB C 5.34 4.28 31.69 D 5.97 ElevD SDDA = 155.3 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” o are in units of feet. [pause] To find the zenith angle D A, --- D o A plan view o o B C

Find: ZenithDA all lengths are in feet SDDA = 155.3 [ft] 89 29’15” BS FS Elev are in feet A 3.27 ElevA B 4.56 4.31 ElevB C 5.34 4.28 31.69 D 5.97 ElevD SDDA = 155.3 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” o we’ll consider a section view showing Point D, ---- D o A plan view o o B C

Find: ZenithDA all lengths are in feet X’ D X A plan A view D section BS FS Elev are in feet A 3.27 ElevA B 4.56 4.31 ElevB C 5.34 4.28 31.69 D 5.97 ElevD X’ D X and Point A. The zenith angle D A is the angle between a vertical line at Point D, ---- A plan A view D section view Section X-X’ B C

Find: ZenithDA ZDA all lengths are in feet X’ D X A plan A view D BS FS Elev are in feet A 3.27 ElevA B 4.56 4.31 ElevB C 5.34 4.28 31.69 D 5.97 ElevD ZDA X’ D X and the line segment connecting Points D and A. To better define this angle, we’ll add a Point A prime, ---- A plan A view D section view Section X-X’ B C

Find: ZenithDA ZDA all lengths are in feet X’ A’ D X A plan A view D BS FS Elev are in feet A 3.27 ElevA B 4.56 4.31 ElevB C 5.34 4.28 31.69 D 5.97 ElevD ZDA X’ A’ D X which is a projection of Point A, on to the vertical line segment through Point D. Now we’ll take a closer look at --- A plan A view D section view Section X-X’ B C

Find: ZenithDA ZDA A A’ D X’ A’ D X A plan A view D section view B C BS FS Elev A 3.27 ElevA B 4.56 4.31 ElevB D C 5.34 4.28 31.69 D 5.97 ElevD ZDA X’ A’ D X Triangle A, A prime, D. [pause] Zenith Angle D A, is sketched in, and we notice, ---- A plan A view D section view Section X-X’ B C

Find: ZenithDA ZDA A A’ D X’ A’ D X A plan A view D section view B C BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D C 5.34 4.28 31.69 D 5.97 ElevD ZDA X’ A’ D X the cosine of Zenith Angle D A equals, the length of line segment --- A plan A view D section view Section X-X’ B C

Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA X’ D X A plan view B C BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 cos (ZDA)= D 5.97 ElevD LDA X’ D X D A prime, divided by length of line segment D A. So the zenith equal ---- A plan view B C

Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA X’ LDA’ D ZDA=cos-1 X A plan BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 cos (ZDA)= D 5.97 ElevD LDA X’ LDA’ D ZDA=cos-1 X the arc cosine of that quotient. The problem statement provides the slope distance D A as --- A plan LDA view B C

Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA X’ LDA’ D ZDA=cos-1 X A plan BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 cos (ZDA)= D 5.97 ElevD LDA X’ LDA’ D ZDA=cos-1 X 155.3 feet, but we don’t know the distance between Points D and A prime. --- A plan LDA view SDDA = 155.3 [ft] B C

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA X’ LDA’ D ZDA=cos-1 X A BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 ElevD LDA X’ LDA’ D ZDA=cos-1 X [pause] To find this length, we’ll turn to the level survey data, --- A plan LDA view SDDA = 155.3 [ft] B C

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA X’ LDA’ D ZDA=cos-1 X A BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 ElevD LDA X’ LDA’ D ZDA=cos-1 X and equate the length of line segment D A prime, to the elevation of Point A prime, minus --- A plan LDA view SDDA = 155.3 [ft] B C

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA’ - ElevD BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 ElevD LDA LDA’ LDA’ = ElevA’ - ElevD ZDA=cos-1 the elevation of Point D. Where the elevation of Point A prime, is equivalent to, the elevation of --- LDA SDDA = 155.3 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA’ - ElevD BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 ElevD LDA LDA’ LDA’ = ElevA’ - ElevD ZDA=cos-1 Point A. The elevations for points A and D can be determined from --- LDA ElevA’ = ElevA SDDA = 155.3 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA’ - ElevD BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 ElevD LDA LDA’ LDA’ = ElevA’ - ElevD ZDA=cos-1 the level survey data. [pause] To find the elevation at Point D, we’ll add the elevation from Point C, --- LDA ElevA’ = ElevA SDDA = 155.3 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 ElevD LDA LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 to the backsight reading at Point C, minus the foresight reading to Point D. After plugging in --- LDA ElevD = ElevC + BSC - FSD SDDA = 155.3 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 ElevD LDA LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 these values, the elevation at Point D equals, --- LDA ElevD = ElevC + BSC - FSD SDDA = 155.3 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 31.06 LDA LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 31.06 feet. The elevation at Point A is equal to the elevation at Point C, --- LDA ElevD = ElevC + BSC - FSD ElevD = 31.06 [ft] SDDA = 155.3 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 31.06 LDA LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 minus the backsight readings to Points A and B, plus the foresight readings to Points B and C. LDA ElevA = ElevC - BSB - BSA + FSB + FSC SDDA = 155.3 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 ElevA B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 31.06 LDA LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 After making these substitutions, the elevation at Point A equals, --- LDA ElevA = ElevC - BSB - BSA + FSB + FSC SDDA = 155.3 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 32.45 B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 31.06 LDA LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 32.45 feet. [pause] LDA ElevA = ElevC - BSB - BSA + FSB + FSC SDDA = 155.3 [ft] ElevA = 32.45 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 32.45 B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 31.06 LDA LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 32.45 feet minus 31.06 feet, equals, --- LDA ElevA = ElevC - BSB - BSA + FSB + FSC SDDA = 155.3 [ft] ElevA = 32.45 [ft]

? Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 32.45 B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 ? cos (ZDA)= D 5.97 31.06 LDA LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 1.39 feet, the length of line segment. [pause] Lastly, we’ll take the arc cosine of the quotient of --- LDA LDA’ = 1.39 [ft] SDDA = 155.3 [ft] ElevA = 32.45 [ft]

Find: ZenithDA A A’ D LDA’ cos (ZDA)= LDA LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 32.45 B 4.56 4.31 ElevB D LDA’ C 5.34 4.28 31.69 cos (ZDA)= D 5.97 31.06 LDA LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 of the two line segments D A prime divided D A, and the Zenith angle D A equals, --- LDA LDA’ = 1.39 [ft] SDDA = 155.3 [ft] ElevA = 32.45 [ft]

Find: ZenithDA A A’ D SDDA = 155.3 [ft] LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 32.45 B 4.56 4.31 ElevB D C 5.34 4.28 31.69 SDDA = 155.3 [ft] D 5.97 31.06 LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 89.48717 degrees. After converting this angle to degree minute second format, we have --- LDA LDA’ = 1.39 [ft] o ZDA=89.48717

Find: ZenithDA A A’ D SDDA = 155.3 [ft] LDA’ LDA’ = ElevA - ElevD BS FS Elev ZDA A 3.27 32.45 B 4.56 4.31 ElevB D C 5.34 4.28 31.69 SDDA = 155.3 [ft] D 5.97 31.06 LDA’ LDA’ = ElevA - ElevD ZDA=cos-1 89 degrees, 29 minutes and 14 seconds. [pause] LDA LDA’ = 1.39 [ft] o ZDA=89.48717 o ZDA=89 29’14”

Find: ZenithDA A A’ D SDDA = 155.3 [ft] LDA’ = 1.39 [ft] 89 29’15” BS FS Elev ZDA A 3.27 32.45 B 4.56 4.31 ElevB D C 5.34 4.28 31.69 SDDA = 155.3 [ft] D 5.97 31.06 LDA’ = 1.39 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” LDA’ o ZDA=cos-1 When reviewing the possible solutions, --- o LDA o o ZDA=89.48717 o o ZDA=89 29’14”

Find: ZenithDA A A’ D SDDA = 155.3 [ft] LDA’ = 1.39 [ft] 89 29’15” BS FS Elev ZDA A 3.27 32.45 B 4.56 4.31 ElevB D C 5.34 4.28 31.69 SDDA = 155.3 [ft] D 5.97 31.06 LDA’ = 1.39 [ft] 89 29’15” 89 46’30” 90 14’30” 90 30’45” LDA’ o ZDA=cos-1 the answer is A. o LDA AnswerA o o ZDA=89.48717 o o ZDA=89 29’14”

Find the average slope from point A, to Point D, as a percent [pause] In this problem, -----