Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 15

Convert the following \(\mathrm{RB}\) to \(\mathrm{WCB}\) (a) \(\mathrm{N} 32^{\circ} 30^{\prime} \mathrm{E}\) (b) \(\mathrm{S} 42^{\circ} 40^{\prime} \mathrm{E}\) (c) \(\mathrm{S} 60^{\circ} 30^{\prime} \mathrm{W}\) (d) \(\mathrm{N} 72^{\circ} 50^{\prime} \mathrm{W}\)

Short Answer

Expert verified
Question: Convert the following Relative Bearings to Whole Circle Bearings: (a) N 32°30’ E (b) S 42°40’ E (c) S 60°30’ W (d) N 72°50’ W Answer: (a) WCB = 32°30' (b) WCB = 137°20' (c) WCB = 240°30' (d) WCB = 287°10'

Step by step solution

01

Case (a): N 32°30’ E

To convert N 32°30’ E to WCB, we start at North and move 32°30’ towards East in a clockwise direction. There's no need to do any calculations in this case, as the WCB is the same as the given RB. So, the WCB is 32°30’.
02

Case (b): S 42°40’ E

To convert S 42°40’ E to WCB, we start at South (which is 180° from North) and move 42°40’ towards East in a clockwise direction. We calculate the WCB as follows: \(\text{WCB} = 180° - 42°40'\) \(\text{WCB} = 137°20'\)
03

Case (c): S 60°30’ W

To convert S 60°30’ W to WCB, we again start at South (180°) and move 60°30’ towards West in a clockwise direction. We calculate the WCB as follows: \(\text{WCB} = 180° + 60°30'\) \(\text{WCB} = 240°30'\)
04

Case (d): N 72°50’ W

ToStr convert N 72°50’ W to WCB, we start at North and move 72°50’ towards West in a clockwise direction. To find the WCB, we subtract the RB from 360 degrees: \(\text{WCB} = 360° - 72°50'\) \(\text{WCB} = 287°10'\) In summary, we have converted the given RB to WCB as follows: (a) \(\mathrm{N} 32^{\circ} 30^{\prime} \mathrm{E}\) → \(\text{WCB} = 32°30'\) (b) \(\mathrm{S} 42^{\circ} 40^{\prime} \mathrm{E}\) → \(\text{WCB} = 137°20'\) (c) \(\mathrm{S} 60^{\circ} 30^{\prime} \mathrm{W}\) → \(\text{WCB} = 240°30'\) (d) \(\mathrm{N} 72^{\circ} 50^{\prime} \mathrm{W}\) → \(\text{WCB} = 287°10'\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ammonium salt of an organic acid (A) is heated to give (B). (B) is treated with bromine and alkali to give (C). (C) with nitrous acid gives ethanol. (A) is (a) ethanoic acid (b) methanoic acid (c) propanoic acid (d) propanone

The structural feature which distinguishes proline from other amino acids is that (a) it is optically inactive (b) it is a secondary amine (c) it contains two carboxylic acid groups (d) it contains two amino groups

Which of the following compounds on oxidation with acidified dichromate gives p-benzoquinone?

Which of the following compounds react with aldehyde to form Schiff's base? (a) \(1^{\circ}\) amines (b) \(2^{\circ}\) amines (c) \(3^{\circ}\) amines (d) amides

Each question in this section has four suggested answers of which ONE OR MORE answers will be correct. Which of the following reactions will give phenyl cyanide as a product? (a) Reaction of benzene diazonium chloride with cuprous cyanide and KCN (b) Heating bromobenzene with alcoholic potassium cyanide (c) Dehydration of benzaldoxime by heating with acetic anhydride (d) Fusing sodium benzene sulphonate with \(\mathrm{NaCN}\)
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Chemical Reactions

Read Explanation

Inorganic Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free