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Chapter 6: Problem 14

Convert the following WCB to \(\mathrm{RB}\) (a) \(66^{\circ} 30^{\prime}\) (b) \(130^{\circ} 15^{\prime}\) (c) \(205^{\circ} 20^{\prime}\) (d) \(265^{\circ} 10^{\prime}\) (e) \(295^{\circ} 30^{\prime}\) (f) \(320^{\circ} 15^{\prime}\)

Short Answer

Expert verified
Question: Convert the following whole circle bearings to reduced bearings: (a) \(66^{\circ} 30^{\prime}\) (b) \(130^{\circ} 15^{\prime}\) (c) \(205^{\circ} 20^{\prime}\) (d) \(265^{\circ} 10^{\prime}\) (e) \(295^{\circ} 30^{\prime}\) (f) \(320^{\circ} 15^{\prime}\) Answer: (a) \(66^{\circ} 30^{\prime}\) NE (b) \(40^{\circ} 15^{\prime}\) SE (c) \(25^{\circ} 20^{\prime}\) SW (d) \(4^{\circ} 50^{\prime}\) NW (e) \(25^{\circ} 30^{\prime}\) NW (f) \(50^{\circ} 15^{\prime}\) NW

Step by step solution

01

Identify the Quadrant for each WCB

Identify the quadrant in which each of the WCB bearings fall into. Recall that the quadrants are: 1. Northeast (NE): \(0^{\circ} < \text{WCB} < 90^{\circ}\) 2. Southeast (SE): \(90^{\circ} < \text{WCB} <180^{\circ}\) 3. Southwest (SW): \(180^{\circ} < \text{WCB} <270^{\circ}\) 4. Northwest (NW): \(270^{\circ} < \text{WCB} <360^{\circ}\)
02

Convert each WCB into a reduced bearing

For each WCB, calculate the reduced bearing based on the slope quadrant identified in the step 1.
03

(a) \(66^{\circ} 30^{\prime}\)

Since \(66^{\circ} 30^{\prime}\) falls between \(0^{\circ}\) and \(90^{\circ}\), it is in the Northeast (NE) quadrant. Therefore, the reduced bearing is the same as the WCB. So, RB = \(66^{\circ} 30^{\prime}\) NE.
04

(b) \(130^{\circ} 15^{\prime}\)

Since \(130^{\circ} 15^{\prime}\) falls between \(90^{\circ}\) and \(180^{\circ}\), it is in the Southeast (SE) quadrant. To find the RB, subtract \(90^{\circ}\) from the WCB: \(130^{\circ} 15^{\prime}\) - \(90^{\circ}\) = \(40^{\circ} 15^{\prime}\). So, RB = \(40^{\circ} 15^{\prime}\) SE.
05

(c) \(205^{\circ} 20^{\prime}\)

Since \(205^{\circ} 20^{\prime}\) falls between \(180^{\circ}\) and \(270^{\circ}\), it is in the Southwest (SW) quadrant. To find the RB, subtract \(180^{\circ}\) from the WCB: \(205^{\circ} 20^{\prime}\) - \(180^{\circ}\) = \(25^{\circ} 20^{\prime}\). So, RB = \(25^{\circ} 20^{\prime}\) SW.
06

(d) \(265^{\circ} 10^{\prime}\)

Since \(265^{\circ} 10^{\prime}\) falls between \(270^{\circ}\) and \(360^{\circ}\), it is in the Northwest (NW) quadrant. To find the RB, subtract \(270^{\circ}\) from the WCB: \(265^{\circ} 10^{\prime}\) - \(270^{\circ}\) = -\(4^{\circ} 50^{\prime}\). So, RB = \(4^{\circ} 50^{\prime}\) NW.
07

(e) \(295^{\circ} 30^{\prime}\)

Since \(295^{\circ} 30^{\prime}\) falls between \(270^{\circ}\) and \(360^{\circ}\), it is in the Northwest (NW) quadrant. To find the RB, subtract \(270^{\circ}\) from the WCB: \(295^{\circ} 30^{\prime}\) - \(270^{\circ}\) = \(25^{\circ} 30^{\prime}\). So, RB = \(25^{\circ} 30^{\prime}\) NW.
08

(f) \(320^{\circ} 15^{\prime}\)

Since \(320^{\circ} 15^{\prime}\) falls between \(270^{\circ}\) and \(360^{\circ}\), it is in the Northwest (NW) quadrant. To find the RB, subtract \(270^{\circ}\) from the WCB: \(320^{\circ} 15^{\prime}\) - \(270^{\circ}\) = \(50^{\circ} 15^{\prime}\). So, RB = \(50^{\circ} 15^{\prime}\) NW.

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