Title MSTE 33 Solutions.pdf ✅

There are two possible answers: 30.2256 m or 14.9810 m . ▣ 4. Find the bearing of line 4-5. [SOLUTION] Solve for A using cosine law At x = 30.2256, A = cos−1 332 + 30.22562 − 39.26592 2(33)(30.2256) = 76°38′43" Bearing angle: 180° − 76°38′43" − 33°09′ = 70°12′17" Bearing: S 70°12′17" W . At x = 14.9810, A = cos−1 332 + 14.98102 − 39.26592 2(33)(14.9810) = 103°21′17" Bearing angle: 180° − 103°21′17" − 33°09′ = 70°12′17" Bearing: S 43°29′43" W . SITUATION 3. Line Bearing Distance (m) 1-2 N 76° E 80.40 2-3 S 36° E 46.30 3-4 S 61° W 72.80 4-5 N 38° W 68.00 ▣ 5. Determine the error of closure. [SOLUTION] Bearing Distance (m) Latitude Departure N 76° E 80.40 19.4505 78.0118 S 36° E 46.30 -37.4575 27.2145 S 61° W 72.80 -35.2941 -63.6723 N 38° W 68.00 53.5847 -41.8650 ∑Lat = 0.2836 ∑Dep = −0.32162
e = √0.28362 + (−0.32162) 2 e = 0.4288 m ▣ 6. Using the compass rule, find the correct bearing of line 2-3. [SOLUTION] ∑Distances = 267.50 Dep2−3 = 27.2145 − 46.30 267.50 (−0.32162) Dep2−3 = 27.27017 Lat2−3 = −37.4575 − 46.30 267.50 (0.2836) Lat2−3 = −37.5066 θ = tan−1 27.27017 37.5066 θ = 36°1′12" Bearing: S 36°1′12" E . ▣ 7. Using compass rule, find the correct distance of line 2-3. [SOLUTION] L2−3 = √(27.27017) 2 + (37.5066) 2 L2−3 = 46.3725 m ▣ 8. Using the transit rule, find the correct bearing of line 2-3. [SOLUTION] Lat2−3 = −37.4575 − 37.4575 145.7868 (0.2836) Lat2−3 = −37.5304 Dep2−3 = 27.2145 − 27.2145 210.7636 (−0.32162) Dep2−3 = 27.2560 θ = tan−1 27.2560 37.5304 θ = 35°59′19" Bearing: S 35°59′19" E .