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 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 = $50\div 80=0.625=tan^{1}(0.625)=32.0=32.0-150=118.$ Last edited by 53453; May 30th, 2014 at 11:05 AM.
 May 30th, 2014, 12:19 PM #2 Global Moderator   Joined: May 2007 Posts: 6,730 Thanks: 689 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: 471 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: 471 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: 471 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. Thanks from 53453
May 30th, 2014, 12:49 PM   #7
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Joined: Jul 2013
From: United Kingdom

Posts: 471
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Quote:
 Originally Posted by 53453 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?
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: 471 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 Thank-you for being so patient... I really fervently appreciate it.

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