May 31st, 2014, 11:12 AM  #11 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 
My pleasure. You know, maths teaches us to be patient, because a lot of things in maths can't be understood right away. If something hasn't sunk in, just keep asking questions and searching for answers. Those who seek shall find. 
May 31st, 2014, 11:28 AM  #12 
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038 
A bearing is usually given as an angle measured clockwise in relation to due north, so the angle shown in the diagram as 150° ought to be shown as 30°.

May 31st, 2014, 12:05 PM  #13 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40  I would agree. The 150 degree angle should be turned into a 30 degree angle to avoid confusion.
Last edited by perfect_world; May 31st, 2014 at 12:09 PM. 
May 31st, 2014, 12:10 PM  #14 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40  therefore this IS the answer, if we are to follow the logic.

May 31st, 2014, 01:16 PM  #15 
Newbie Joined: May 2014 From: England Posts: 16 Thanks: 0 
Do you mean something like this? Is the cos 150 or 30? 
May 31st, 2014, 01:26 PM  #16 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 
Yep. That's the one. If you're using Microsoft paint, don't forget to label the distances and angles using ordinary text.

June 1st, 2014, 10:01 AM  #17 
Newbie Joined: May 2014 From: England Posts: 16 Thanks: 0 
Using cos(30), have I managed to correctly use the Sine Rule to find B? \[\frac{a}{sin}= \frac{b}{sin}=\frac{44.4}{sin30}=\frac{80}{sinb}\] \[\frac{80\cdot sin30}{44.4}=0.9009009009\] \[sin^{1}(0.9009009009)= 64.27674048\] \[B=64.28\] 
June 4th, 2014, 11:01 PM  #18 
Newbie Joined: Aug 2013 From: UK Posts: 5 Thanks: 0 
good

June 5th, 2014, 12:58 AM  #19  
Global Moderator Joined: Dec 2006 Posts: 20,469 Thanks: 2038  Quote:
Also, angle B is obtuse, so you need to subtract the value you obtained from 180 to get the correct number of degrees for B. In your original post, $50\div 80=0.625=\tan^{1}(0.625)$ was incorrect on two counts: (1) The method was incorrect  you apparently guessed 50/80 for the angle's tangent, and (2) The inverse for tan is denoted by $\tan^{1}$, not $\tan^1$. Last edited by skipjack; June 5th, 2014 at 01:09 AM.  

Tags 
bearings 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Trigonometry and bearings  sakuraa  Trigonometry  6  February 25th, 2013 08:49 AM 
Bearings  taylor_1989_2012  Algebra  5  September 11th, 2012 11:49 PM 
Trig. Bearings  Sunyata.  Trigonometry  1  October 29th, 2011 01:28 AM 
Help on bearings  a.sundar23  Algebra  1  May 5th, 2009 06:46 PM 
Pytragoras and Bearings need help  SuperMaths  Algebra  1  January 12th, 2009 01:02 PM 