May 30th, 2014, 10:53 AM  #1 
Newbie Joined: May 2014 From: England Posts: 16 Thanks: 0  Bearings.
A question regarding bearing and distance.. could someone tell me if my calculations are wrong and if so how? (1) B is an outlet from a Nuclear fuel reprocessing unit which discharges waste at sea. A conservation group wishes to monitor the radioactivity. They place a buoy (A) 50m due north of (B) and another buoy (C) at a bearing of 150 from (A). A line with detectors on it is used between (A) and (C) and this line is 80m long. Calculate the distance BC to 3 s.f. and the bearing of C from B to the nearest degree. distance = \[d^{2}=80^{2} + 50^{2}= d^{2}=6400 + 2500=8900 =\sqrt{8900}=94.3\] Bearing = \[50\div 80=0.625=tan^{1}(0.625)=32.0=32.0150=118.\] Last edited by 53453; May 30th, 2014 at 11:05 AM. 
May 30th, 2014, 12:19 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,524 Thanks: 588 
Your picture is thoroughly confusing. 150 deg is not an acute angle.

May 30th, 2014, 12:28 PM  #3 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 
We sometimes get similar diagrams in examination papers. However, we are told in advance that some diagrams aren't completely accurate. The so called '150' degree angle does look funny I have to admit. 
May 30th, 2014, 12:38 PM  #4 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 
Right, you shouldn't be using Pythagoras's formula to find the distance BC. What you should know: Note that these are small letters. a=BC b=80 c=50 Angle at A=150 degrees. To find the length of BC, which is a (small), use the Cosine Formula: $\displaystyle { a }^{ 2 }={ b }^{ 2 }+{ c }^{ 2 }2bc\cdot cosA$ which is the same as: $\displaystyle { BC }^{ 2 }={ b }^{ 2 }+{ c }^{ 2 }2bc\cdot cosA$ 
May 30th, 2014, 12:40 PM  #5 
Newbie Joined: May 2014 From: England Posts: 16 Thanks: 0 
I should really say the image is not drawn to scale (or accurately). But I thought the principle information regarding 150 degrees and both measurement (50m and 80m) would suffice?

May 30th, 2014, 12:45 PM  #6 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 
Once you know the distance BC, you can use the Sine Rule to get the bearing of C from B. $\displaystyle \frac { a }{ sinA } =\frac { b }{ sinB } =\frac { c }{ sinC } $ If you use the Sine Rule correctly, you'll manage to find all the other angles, and presto  the bearing shall be revealed. 
May 30th, 2014, 12:49 PM  #7 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40  It does. Just refresh the page, as I've given you the tools you'll need to solve the problem, i.e the Cosine Rule and the Sine Rule.

May 31st, 2014, 08:21 AM  #8 
Newbie Joined: May 2014 From: England Posts: 16 Thanks: 0 
Are my calculations correct? \[a^{2}=b^{2} + c^{2}  2\cdot b\cdot c\cdot cosA\] \[50^{^{2}}+80^{2}=8900\] \[2\cdot 50\cdot 80\cdot cos(30)=6928.20323\] \[8900  6928.20323 = 1971.79677\] \[\sqrt{1971.79677} = 44.4049183087\] Last edited by 53453; May 31st, 2014 at 08:25 AM. 
May 31st, 2014, 09:40 AM  #9 
Senior Member Joined: Jul 2013 From: United Kingdom Posts: 468 Thanks: 40 
Nope. The length BC has to be the longest length on the triangle as angle A is 150 degrees. Think about that for one moment. Big A=150 degrees. $\displaystyle { a }^{ 2 }={ BC }^{ 2 }={ 80 }^{ 2 }+{ 50 }^{ 2 }2\left( 80 \right) \left( 50 \right) \cdot cos{ 150 }^{ \circ }\\ \\ { a }^{ 2 }={ BC }^{ 2 }=15828.20323\\ \\ \therefore \quad a=BC=\sqrt { 15828.20323 } =12{ 6 }m\quad \left( to\quad the\quad nearest\quad metre \right) $ Last edited by perfect_world; May 31st, 2014 at 09:47 AM. 
May 31st, 2014, 10:33 AM  #10 
Newbie Joined: May 2014 From: England Posts: 16 Thanks: 0 
Thankyou for being so patient... I really fervently appreciate it.


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