Technological University of the Philippines Ayala Blvd., corner San Marcelino St., Ermita Manila
College of Engineering Civil Engineering Department
CE 25-3D
Elementary and Higher Surveying (Lec)
Assignment No.2
Theory of Errors and Measurements
Marquez, Jhyrelle A. 14-205-204
Date of Submission: January 6, 2015
Engr. Jesus Ray M. Mansayon Instructor
1. Given the dimensions of the following tracts of land: a.) 108.75 m by 76.82 m b.) 940.08 m by 1296.73 m c.) 13.36 m by 50.08 m d.) 1258.30 m by 624.03 m e.) 8476.55 m by 121.79 m
Determine the area of each tract in square meters, square kilometers, ares, and hectares. Given: a.) 108.75 m by 76.82 m b.) 940.08 m by 1296.73 m c.) 13.36 m by 50.08 m d.) 1258.30 m by 624.03 m e.) 8476.55 m by 121.79 m Required:
area in square meter area in square kilometer area in ares area in hectares Solution: 𝐴𝑟𝑒𝑎 = 𝐿𝑒𝑛𝑔𝑡ℎ × 𝑊𝑖𝑑𝑡ℎ a.) 𝐴 = 108.75𝑚2 × 76.82𝑚2 = 𝟖𝟑𝟓𝟒. 𝟏𝟕𝟓 𝒎𝟐 𝐴 = 8354.175 𝑚2 × 1 𝑘𝑚2 1000000 𝑚2 = 𝟎. 𝟎𝟎𝟖𝟑𝟓𝟒 𝒌𝒎𝟐 𝐴 = 8354.175 𝑚2 × 1 𝑎𝑟𝑒 100 𝑚2 = 𝟖𝟑. 𝟓𝟒𝟐 𝒂𝒓𝒆𝒔 𝐴 = 83.542 𝑎𝑟𝑒𝑠 × 1 ℎ𝑒𝑐𝑡𝑎𝑟𝑒100 𝑎𝑟𝑒𝑠 = 𝟎. 𝟖𝟑𝟓𝟒 𝒉𝒆𝒄𝒕𝒂𝒓𝒆𝒔
b.) 𝐴 = 940.08𝑚 × 1296.73𝑚 = 𝟏𝟐𝟏𝟗𝟎𝟐𝟗. 𝟗𝟒 𝒎𝟐 𝐴 = 1219029.94 𝑚2× 1 𝑘𝑚2 1000000 𝑚2 = 𝟏. 𝟐𝟏𝟗𝟎𝟑 𝒌𝒎𝟐 𝐴 = 1219029.94 𝑚2× 1 𝑎𝑟𝑒 100 𝑚2 = 𝟏𝟐𝟏𝟗𝟎. 𝟑𝟎 𝒂𝒓𝒆𝒔 𝐴 = 12190.30 𝑎𝑟𝑒𝑠 × 1 ℎ𝑒𝑐𝑡𝑎𝑟𝑒100 𝑎𝑟𝑒𝑠 = 𝟏𝟐𝟏. 𝟗𝟎𝟑 𝒉𝒆𝒄𝒕𝒂𝒓𝒆𝒔 c.) 𝐴 = 13.36𝑚 × 50.08𝑚 = 𝟔𝟔𝟗. 𝟎𝟔𝟖𝟖 𝒎𝟐 𝐴 = 669.0688 𝑚2× 1 𝑘𝑚2 1000000 𝑚2 = 𝟎. 𝟎𝟎𝟎𝟔𝟔𝟗 𝒌𝒎𝟐 𝐴 = 669.0688 𝑚2× 1 𝑎𝑟𝑒 100 𝑚2 = 𝟔. 𝟔𝟗𝟎𝟔𝟖𝟖 𝒂𝒓𝒆𝒔 𝐴 = 6.690688 𝑎𝑟𝑒𝑠 × 1 ℎ𝑒𝑐𝑡𝑎𝑟𝑒100 𝑎𝑟𝑒𝑠 = 𝟎. 𝟎𝟔𝟔𝟗𝟎𝟕 𝒉𝒆𝒄𝒕𝒂𝒓𝒆𝒔
d.) 𝐴 = 1258.30𝑚 × 624.03𝑚 = 𝟕𝟖𝟓𝟐𝟏𝟔. 𝟗𝟒𝟗𝒎𝟐 𝐴 = 785216.949 𝑚2× 1 𝑘𝑚2 1000000 𝑚2 = 𝟎. 𝟕𝟖𝟓𝟐𝟏𝟕 𝒌𝒎𝟐 𝐴 = 785216.949 𝑚2× 1 𝑎𝑟𝑒 100 𝑚2 = 𝟕𝟖𝟓𝟐. 𝟏𝟔𝟗 𝒂𝒓𝒆𝒔 𝐴 = 7852.169 𝑎𝑟𝑒𝑠 × 1 ℎ𝑒𝑐𝑡𝑎𝑟𝑒100 𝑎𝑟𝑒𝑠 = 𝟕𝟖. 𝟓𝟐𝟏𝟔𝟗𝟒𝟗 𝒉𝒆𝒄𝒕𝒂𝒓𝒆𝒔 e.) 𝐴 = 8476.55𝑚 × 195.42𝑚 = 𝟏𝟎𝟑𝟐𝟑𝟓𝟗. 𝟎𝟐𝒎𝟐 𝐴 = 1032359.02 𝑚2× 1 𝑘𝑚2 1000000 𝑚2 = 𝟏. 𝟎𝟑𝟐𝟑𝟓𝟗 𝒌𝒎𝟐 𝐴 = 1032359.02 𝑚2× 1 𝑎𝑟𝑒 100 𝑚2 = 𝟏𝟎𝟑𝟐𝟑. 𝟓𝟗 𝒂𝒓𝒆𝒔 𝐴 = 10323.59 𝑎𝑟𝑒𝑠 × 1 ℎ𝑒𝑐𝑡𝑎𝑟𝑒100 𝑎𝑟𝑒𝑠 = 𝟏𝟎𝟑. 𝟐𝟑𝟓𝟗 𝒉𝒆𝒄𝒕𝒂𝒓𝒆𝒔
2. Following are the dimensions for length, width, and depth of five excavated borrow pits for a highway project.
a.) 113.26 m, 35.48 m, and 18.60 m b.) 50.08 m, 39.25 m, and 7.14 m c.) 243.55 m, 76.19 m, and 24.66 m d.) 42.055 m, 8.605 m, and 12.332 m e.) 9.5 m, 6.3 m, and 4.9 m
Determine the volume of each pit in cubic meters.
Given: a.) 113.26 m, 35.48 m, and 18.60 m b.) 50.08 m, 39.25 m, and 7.14 m c.) 243.55 m, 76.19 m, and 24.66 m d.) 42.055 m, 8.605 m, and 12.332 m e.) 9.5 m, 6.3 m, and 4.9 m Required:
Volume in cubic meters. Illustration: Solution: 𝑉𝑜𝑙𝑢𝑚𝑒 = 𝐿𝑒𝑛𝑔𝑡ℎ × 𝑊𝑖𝑑𝑡ℎ × 𝐻𝑒𝑖𝑔ℎ𝑡 a.) 𝑉 = 113.26𝑚 × 35.48𝑚 × 18.60𝑚 = 𝟕𝟒𝟕𝟒𝟑. 𝟒𝟒𝟓𝟐𝟖 𝒎𝟑 b.) 𝑉 = 50.08𝑚 × 39.25𝑚 × 7.14𝑚 = 𝟏𝟒𝟎𝟑𝟒. 𝟔𝟔𝟗𝟔 𝒎𝟑 c.) 𝑉 = 243.55𝑚 × 76.19𝑚 × 24.66𝑚 = 𝟒𝟓𝟕𝟓𝟗𝟐. 𝟕𝟗𝟕𝟐 𝒎𝟑 d.) 𝑉 = 42.055𝑚 × 8.605𝑚 × 12.332𝑚 = 𝟒𝟒𝟔𝟐. 𝟕𝟒𝟒𝟓𝟒𝟕 𝒎𝟑 e.) 𝑉 = 9.5𝑚 × 6.3𝑚 × 4.9𝑚 = 𝟐𝟗𝟑. 𝟐𝟔𝟓 𝒎𝟑
3. Given the following numbers: 45.63, 5.700, 4010, 0.00037, 0.000940, 6.0090, 7.00, 9.5 x 108, 4.00 x 107, 2.604 x 1018 and 3.00 x 10-16. For each number, identify the
significant figures. Tabulate values accordingly. Given: 45.63 0.000940 4.00 x 107 5.700 6.0090 2.604 x 1018 4010 7.00 3.00 x 10-16 0.00037 9.5 x 108 Required:
significant figures of each number Illustration: Solution: NUMBER SIGNIFICANT FIGURES NUMBER OF SIGNIFICANT FIGURES 45.63 4, 5, 6,3 4 5.700 5, 7, 0 4 4010 4, 1, 0 4 0.00037 3, 7 2 0.000940 9, 4, 0 3 6.0090 6, 0, 9 5 7.00 7, 0 3 9.5 x 108 9, 5 2 4.00 x 107 4, 0 3 2.604 x 1018 2, 6, 0, 4 4 3.00 x 10-16 3, 0 3
42˚ 05ʹ
115˚ 38ʹ
22˚ 08ʹ
4. The three angles of a triangle were measured with the following results: A = 42˚ 05ʹ, B = 115˚ 38ʹ and C = 22˚ 08ʹ. Determine the most probable value of each angle.
Given:
A = 42˚ 05ʹ B = 115˚ 38ʹ C = 22˚ 08ʹ
Required:
Most probable value of each angle (xˉ ) Illustration: B A C Solution: ∠ x e c xˉ = 𝒙 + 𝒄 A 42˚ 05ʹ 𝑒 = 𝛴𝑥 − 𝑇𝑣 = 179˚ 51’ − 180˚ = +𝟗′ 9′× 1 3= +3′ 𝟒𝟐˚ 𝟎𝟖’ B 115˚ 38ʹ 9′× 1 3= +3′ 𝟏𝟏𝟓˚ 𝟒𝟏’ C 22˚ 08ʹ 9′× 1 3= +3′ 𝟐𝟐˚ 𝟏𝟏’ Total 179˚ 51’
118˚ 44ʹ 15ʺ 100˚ 35ʹ 40ʺ 80˚ 54ʹ 35ʺ 59˚ 45ʹ 50ʺ A B C D
5. The interior angles of a quadrilateral were observed to be: A = 100˚ 35ʹ 40ʺ, B = 118˚ 44ʹ 15ʺ, C = 80˚ 54ʹ 35ʺ, and D = 59˚ 45ʹ 50ʺ. Determine the most probable value of each of these angles.
Given: A = 100˚ 35ʹ 40ʺ B = 118˚ 44ʹ 15ʺ C = 80˚ 54ʹ 35ʺ D = 59˚ 45ʹ 50ʺ Required:
Most probable value of each angle (xˉ ) Illustration: Solution: ∠ x e c xˉ = 𝒙 + 𝒄 A 100˚ 35ʹ 40ʺ 𝑒 = 𝛴𝑥 − 𝑇𝑣 = 360˚ 00’ 20" − 360˚ = −𝟐𝟎" 20"× 1 4= −5" 𝟏𝟎𝟎˚ 𝟑𝟓’ 𝟑𝟓” B 118˚ 44ʹ 15ʺ 20"× 1 4= −5" 𝟏𝟏𝟖˚ 𝟒𝟒’ 𝟏𝟎” C 80˚ 54ʹ 35ʺ 20"× 1 4= −5" 𝟖𝟎˚ 𝟓𝟒’ 𝟑𝟎” D 59˚ 45ʹ 50ʺ 20"× 1 4= −5" 𝟓𝟗˚ 𝟒𝟓’ 𝟒𝟓” Total: 360˚ 00’ 20”
Pt. 1 Pt. 2 x (m)
𝛴𝑥 = 9210.3 𝛴(𝑥 − xˉ )2 = 4.076
𝑛 = 10
6. A surveying instructor sent all the 40 students in his class out to measure a distance between two points marked on a runway. The students working in groups of four came up with 10 different measurements as follows: 920.45, 921.05, 921.65, 920.25, 920.15, 921.85, 921.95, 920.45, 921.15, and 921.35 meters. Assuming these values are equally reliable and that variations result only from accidental errors, determine the relative precision of a single measurement and the relative precision of the mean. Given: 920.45 920.15 921.15 921.05 921.85 921.35 921.65 921.95 920.25 920.45 Required:
relative precision of a single measurement (𝑅𝑃𝑠) relative precision of the mean (𝑅𝑃𝑚)
Illustration: Solution: 𝒏 𝒙 (𝒎) xˉ (𝒎) (𝑥 − xˉ ) 2 (𝑚2) 1 920.45 xˉ = 𝟗𝟐𝟏. 𝟎𝟑 𝒎 0.3364 2 921.05 0.0004 3 921.65 0.3844 4 920.25 0.6084 5 920.15 0.7744 6 921.85 0.6724 7 921.95 0.8464 8 920.45 0.3364 9 921.15 0.0144 10 921.35 0.1024
Computation xˉ =𝜮𝒙𝒏 xˉ =𝟗𝟐𝟏𝟎.𝟑𝟏𝟎 xˉ = 𝟗𝟐𝟏. 𝟎𝟑 𝒎 𝑣 =𝛴(𝑥−𝑛−1xˉ )2 𝑣 =4.07610−1 𝑣 = 𝟎. 𝟒𝟓𝟐𝟖𝟖𝟖𝟖𝟖𝟖𝟗 𝒎𝟐 𝜎 = ±√𝑣 𝜎𝑚= ± 𝜎 √𝑛 𝜎 = ±√0.4528888889 𝑚2 𝜎 𝑚= ±0.0629701991√10 𝜎 = ±𝟎. 𝟎𝟔𝟐𝟗𝟕𝟎𝟏𝟗𝟗𝟏 𝒎 𝜎 = ±𝟎. 𝟐𝟏𝟐𝟖𝟏𝟏𝟖𝟔𝟐𝟕 𝒎 𝑃𝐸𝑠 = ±0.6745 𝜎 𝑃𝐸𝑚 = ±0.6745 𝜎𝑚 𝑃𝐸𝑠 = ±0.6745 (0.0629701991) 𝑃𝐸𝑚 = ±0.6745 (0.2128118627) 𝑃𝐸𝑠 = ±0.4539183993 𝑚 𝑃𝐸𝑚 = ±0.01435416014 𝑚 𝑅𝑃𝑠 = 𝑃𝐸𝑠 xˉ 𝑅𝑃𝑚 = 𝑃𝐸𝑚 xˉ 𝑅𝑃𝑠 = 0.4539183993921.03 𝑅𝑃𝑚 = 0.1435416914921.03 𝑹𝑷𝒔 = 𝟐𝟎𝟐𝟗.𝟎𝟕𝟏 , 𝒔𝒂𝒚 𝟐𝟎𝟎𝟎𝟏 𝑹𝑷𝒔 = 𝟔𝟒𝟏𝟔.𝟒𝟕𝟏 , 𝒔𝒂𝒚 𝟔𝟒𝟎𝟎𝟏
Pt. 2 Pt. 2
Pt. 1 Pt. 1
338.65m (Windy day) 338.37m (calm day)
7. A line is measured on a windy day as 338.65 m. the same line measured 338.37 m on a calm day. If the latter measurement is given four times the reliability of the first, determine the most probable value of the measured line.
Given:
338.65 m - windy day (Weight – 1) 338.37 m - calm day (Weight – 4) Required:
Most probable value (xˉ ) Illustration: Solution: Weather (𝑚) 𝑥 Weight 𝑊𝑒𝑖𝑔ℎ𝑡 × (𝑚) x Windy 338.65 1 338.65 Calm 338.37 4 1353.48 Total 5 1692.13 xˉ =𝜮𝒙𝒏 xˉ =𝟏𝟔𝟗𝟐.𝟏𝟑𝟓 xˉ = 𝟑𝟑𝟖. 𝟒𝟐𝟔𝒎
A
B
C x
8. An angle ABC is measured at different times various instruments and procedures. The results which are assigned certain weights, are as follows: 75˚ 09ʹ 26ʺ, weight of 4; 75˚ 09ʹ 25ʺ, weight of 3; 75˚ 09ʹ 27ʺ, weight of 1. Determine the most probable value of the angle measured.
Given: ∠ x weight A 75˚ 09ʹ 26ʺ 4 B 75˚ 09ʹ 25ʺ 3 C 75˚ 09ʹ 27ʺ 1 Required:
Most probable value of angle (xˉ ) Illustration: Solution ∠ (𝑚) 𝑥 Weight 𝑊𝑒𝑖𝑔ℎ𝑡 × (𝑚) x A 75˚ 09ʹ 26ʺ 4 300˚ 37ʹ 44ʺ B 75˚ 09ʹ 25ʺ 3 225˚ 28ʹ 15ʺ C 75˚ 09ʹ 27ʺ 1 75˚ 09ʹ 27ʺ Total 601˚ 15ʹ 26ʺ 8 601˚ 15ʹ 26ʺ xˉ =𝜮𝒏 xˉ =601˚ 15ʹ 26ʺ𝟖 xˉ = 𝟕𝟓˚ 𝟎𝟗ʹ 𝟐𝟓. 𝟕𝟓ʺ
A
B
C b = 564.75 ± 0.06 m
C = 57˚ 15ʹ 45ʺ
9. Two sides and the included angle of a triangle were measured and the probable error of each value were computed as follows: a = 267.55 m ± 0.05 m, b = 564.75 m ± 0.06 m, and angle C = 57˚ 15ʹ 45ʺ. Determine the area of the triangle and the probable error of the area. Given: a = 267.55 m 𝑃𝐸𝐴 = ± 0.05 m b = 564.75 m 𝑃𝐸𝑏= ± 0.06 m C = 57˚ 15ʹ 45ʺ Required: Area (A)
Probable Error of Area (𝑃𝐸𝐴)
Illustration: Solution: 𝐴 =12𝑎𝑏 sin 𝐶 𝐴 =12(267.55)(564.75) sin( 57˚ 15ʹ 45ʺ) 𝑨 = 𝟔𝟑𝟓𝟒𝟖. 𝟗𝟑𝟒𝟏 𝑃𝐸𝐴 = ±𝑘√(𝑃𝐸) 𝑃𝐸𝐴 = ±12sin 𝐶 √(𝑎 × 𝑃𝐸𝑏)2+ (𝑏 × 𝑃𝐸 𝑎)2 𝑃𝐸𝐴 = ±12sin(57˚ 15ʹ 45ʺ)√(267.55 × 0.05)2+ (564.75 × 0.06)2 𝑷𝑬𝑨= ±𝟏𝟑. 𝟔𝟔𝟏𝟏 𝒎𝟐
C E B
A D
134.10
± 0.040 m 320.63 ± 0.055 m 173.73 ± 0.056 m 160.85 ± 0.050 m
10. A line AE is divided into segments for measurement with a tape. The result were AB = 134.10 m ± 0.040 m, BC = 320.63 m ± 0.055 m, CD = 173.73 m ± 0.056 m, and DE = 160.85 m ± 0.050 m. Determine the length of the line and the probable error of the measured length.
Given: AB = 134.10 m; PEAB = ± 0.040 m BC = 320.63 m; PEBC = ± 0.055 m CD = 173.73 m; PECD = ± 0.056 m DE = 160.85 m; PEDE = ± 0.050 m Required: Length (L)
Probable Error of the Length (PEL)
Illustration: Solution: 𝐿 = 𝐴𝐵 + 𝐵𝐶 + 𝐶𝐷 + 𝐷𝐸 𝐿 = 134.10 + 320.63 + 173.73 + 160.85 𝑳 = 𝟕𝟖𝟗. 𝟑𝟏 𝒎 𝑃𝐸𝐿 = ±√(𝑃𝐸𝐴𝐵)2+ (𝑃𝐸 𝐵𝐶)2+ (𝑃𝐸𝐶𝐷)2+ (𝑃𝐸𝐷𝐸)2 𝑃𝐸𝐿 = ±√(0.040)2+ (0.055)2+ (0.056)2+ (0.050)2 𝑷𝑬𝑳 = ±𝟎. 𝟎𝟏 𝒎
11. The four approximately equal sides of a tract of land were measured and the measurements included the following errors: ± 0.085 m, ± 0.014 m, ± 0.0175 m, and ± 0.205 m, respectively. Determine the probable error for the total length (or perimeter) of the tract. Given: PE1 = ± 0.085 m PE2 = ± 0.014 m PE3 = ± 0.0175 m PE4 = ± 0.205 m Required:
Probable Error (PEP)
Illustration: Solution: 𝑃𝐸𝑃 = ±√(𝑃𝐸1)2+ (𝑃𝐸 2)2+ (𝑃𝐸3)2+ (𝑃𝐸4)2 𝑃𝐸𝑃 = ±√(0.085 )2+ (0.014)2+ (0.0175)2+ (0.205)2 𝑷𝑬𝑷 = ±𝟎. 𝟐𝟖𝟑 𝒎
L = 226.25 ± 0.03 m W = 3 0 7 .2 8 ± 0 .0 4 m
12. Two sides of rectangle were measured being 226.25 m ± 0.03 m and 307.28 m ± 0.04 m. Determine the area of the figure and the probable error of the area. Given:
L = 226.25 m; PEL = ± 0.03 m
W = 307.28 m; PEW = ± 0.04 m
Required: Area (A)
Probable Error of the Area (PEA)
Illustration: Solution: 𝐴 = 𝐿 × 𝑊 𝑃𝐸𝐴= ±√(𝐿 × 𝑃𝐸𝑊)2+ (𝑊 × 𝑃𝐸𝐿)2 𝐴 = 226.25 × 307.28 𝑃𝐸𝐴= ±√(226.2 × 0.04)2+ (307.28 × 0.03)2 𝑨 = 𝟔𝟗𝟓𝟐𝟐. 𝟏𝟎 𝒎𝟐 𝑷𝑬 𝑨 = ±𝟏𝟐. 𝟗𝟐 𝒎𝟐
13. Plane surveying is that type of surveying in which the earth is considered to be a flat surface. Distances and areas involved are of limited extent and the
a.) approximate shape of the earth is considered b.) theoretical shape of the geoid is evaluated c.) exact shape of the earth is disregarded d.) shape of the earth is considered as geoid
Plane surveying is the survey in which the earth surface is assumed to be plane and the curvature of the earth is ignored. The correct answer is c.
14. A photogrammetric survey makes use of photographs taken with specially designed cameras either from
a.) a ship or an elevated ground station b.) the ground surface or underwater c.) a field or a laboratory environment d.) a map or a scaled drawing
e.) airplanes or ground stations
Photogrammetric survey is a type of survey which makes use of photographs taken with specially designed cameras either from airplanes or ground stations. The answer is e.
15. The meter is now defined as a length equal to a.) 1/10,000,000 of the earth’s meridional quadrant
b.) 650,736.37 wavelengths of the bright-red light produced by burning silver electrodes c.) 39.27 inches
d.) 0.001 kilometers
e.) 1,650,763.73 wavelengths of the orange-red light produced by burning krypton at a specified energy level in the spectrum.
The meter is now defined as a length equal to 1,650,763.73 wavelengths of the orange-red light produced by burning krypton at a specified energy level in the spectrum. The answer is e.
16. The sexagesimal units of angular measurement are the a.) grad, centesimal minute, and centesimal second
b.) degree, minute, and second c.) radian and steradian
d.) mil, grad, and radian
e.) hours, minutes, and seconds
Degree, minute, and second are the sexagesimal units of angular measurement. The answer is b.
17. When rounded off to the nearest hundredth, 36.24445 becomes a.) 36.25 b.) 36.26 c.) 36.2 d.) 36.3 e.) 36.24 Answer: b
18. A line, known to be 150.000 m long, is measured five times with a steel tape in the following order: 150.004, 149.998, 149.997, 150.005, and 149.996 meters, respectively. The more accurate of the five measurements is the
a.) 1st measurement b.) 2nd measurement c.) 3rd measurement d.) 4th measurement e.) 5th measurement Answer: b
19. If for particular measurement the probable error of the mean is 0.09 m and the most probable value of the measurement is 362.70 m, the relative precision would be expressed as a.) 1/4030 b.) 0.000248 c.) 1:362.70 d.) 1:0.09 e.) 1/363 Given: 𝑃𝐸𝑚 = ± 0.09 𝑚 𝑥̅ = 362.70 𝑚 Required: Relative Precision (RP) Solution: 𝑅𝑃 =𝑃𝐸𝑚 𝑥̅ 𝑅𝑃 = 362.700.09 𝑅𝑃 = 4030 1 Answer: letter a
A E F B C D 122˚ 31ʹ 02ʺ 123˚ 26ʹ 17ʺ 62˚ 53ʹ 07ʺ 120˚ 15ʹ 47ʺ 130˚ 05ʹ 07ʺ 160˚ 50ʹ 35ʺ
20. The interior angles of a hexagon were observed and recorded as follows: A = 122˚ 31ʹ 02ʺ, B = 123˚ 26ʹ 17ʺ, C = 130˚ 05ʹ 07ʺ, D = 120˚ 15ʹ 47ʺ, E = 160˚ 50ʹ 35ʺ, and F = 62˚ 53ʹ 07ʺ. The discrepancy of the measurement is
a.) 0ʹ 30ʺ b.) 2ʹ 30ʺ c.) 1ʹ 00ʺ d.) 2ʹ 50ʺ e.) 1ʹ 30ʺ Given: A = 122˚ 31ʹ 02ʺ B = 123˚ 26ʹ 17ʺ C = 130˚ 05ʹ 07ʺ D = 120˚ 15ʹ 47ʺ E = 160˚ 50ʹ 35ʺ F = 62˚ 53ʹ 07ʺ Required: Discrepancy (d) Illustration: Solution: ∠ x A 122˚ 31ʹ 02ʺ B 123˚ 26ʹ 17ʺ C 130˚ 05ʹ 07ʺ D 120˚ 15ʹ 47ʺ E 160˚ 50ʹ 35ʺ F 62˚ 53ʹ 07ʺ TOTAL 720˚ 01ʹ 55ʺ
Computation 𝑛 = 6 ∠𝑖𝑇 = (𝑛 − 2)180˚ ∠𝑖𝑇 = (6 − 2)180˚ ∠𝑖𝑇 = 720˚ 𝑑 = |720˚ 01′55"| 𝒅 = 𝟎𝟏′𝟓𝟓"
21. Five measurements were made to determine the length of a line and recorded as follows: 350.33, 350.22, 350.30, 350.27, 350.30 meters. If these measurements were given weights of 4, 5, 1, 4, and 6, respectively, the most probable value of the length measured is a.) 350.26 m b.) 350.29 m c.) 350. 30 m d.) 350.27 m e.) 350.28 m Given: 350.33 m (weight = 4) 350.22 m (weight = 5) 350.30 m (weight = 1) 350.27 m (weight = 4) 350.30 m (weight = 6) Required:
Pt. 1 Pt. 2 x Illustration: Solution 𝑥 (𝑚) Weight 𝑊𝑒𝑖𝑔ℎ𝑡 × x (𝑚) 350.33 4 1401.32 350.22 5 1751.10 350.30 1 350.30 350.27 4 1401.08 350.30 6 2101.80 TOTAL 20 7005.60 𝑥̅ = 𝛴(𝑊×𝑥)𝑊 𝑥̅ = 7005.6020 𝑥̅ = 350.28 𝑚 Answer: letter e
6 9 .4 0 ± 0 .1 6 m 215.50 ± 0.18 m
22. The base and altitude of a triangular lot were measured with certain estimated probable errors as follows: b = 215.50 ± 0.18 m and h = 69.40 ± 0.16 m. The true area of the lot probably
a.) is equal to 7514.52 sq m b.) is equal to 7441.18 sq m
c.) falls between 7441.18 and 7477.85 sq m d.) falls between 7477.85 and 7514.52 sq m e.) falls between 7441.18 and 7514.52 sq m
Given:
b = 215.50 m; PEb = ± 0.18 m
h = 69.40 m; PEh = ± 0.16 m
Required:
True Area of the Lot (A)
Illustration:
Solution:
𝐴 =12𝑏ℎ 𝑃𝐸𝐴= ±12√(215.50 × 0.16)2+ (69.40 × 0.18)2
𝐴 =12(215.50)(69.40) 𝑃𝐸𝐴= ±18.34 𝑚2
𝐴 = 7477.85 𝑚2
Answer: The Area of the lot is 𝟕𝟒𝟕𝟕. 𝟖𝟓 ± 𝟏𝟖. 𝟑𝟒 𝒎𝟐 and it falls between 7451.51
