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May 31st, 2014, 11:12 AM   #11
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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.
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May 31st, 2014, 11:28 AM   #12
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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°.
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May 31st, 2014, 12:05 PM   #13
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Quote:
Originally Posted by skipjack View Post
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°.
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.
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May 31st, 2014, 12:10 PM   #14
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Are my calculations correct?

\[a^{2}=b^{2} + c^{2} - 2\cdot b\cdot c\cdot cosA\]
\[50^{^{2}}+80^{2}=8900\]
\[2\cdot 50\cdot 80\cdot cos(30)=6928.20323\]
\[8900 - 6928.20323 = 1971.79677\]
\[\sqrt{1971.79677} = 44.4049183087\]
therefore this IS the answer, if we are to follow the logic.
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May 31st, 2014, 01:16 PM   #15
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Do you mean something like this?

Is the cos 150 or 30?

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May 31st, 2014, 01:26 PM   #16
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Yep. That's the one. If you're using Microsoft paint, don't forget to label the distances and angles using ordinary text.
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June 1st, 2014, 10:01 AM   #17
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Using cos(30), have I managed to correctly use the Sine Rule to find B?

\[\frac{a}{sin}= \frac{b}{sin}=\frac{44.4}{sin30}=\frac{80}{sinb}\]

\[\frac{80\cdot sin30}{44.4}=0.9009009009\]

\[sin^{1}(0.9009009009)= 64.27674048\]

\[B=64.28\]
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June 4th, 2014, 11:01 PM   #18
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good
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June 5th, 2014, 12:58 AM   #19
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Quote:
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Using cos(30), have I managed to correctly use the Sine Rule to find B?
No. Due to 44.4 having been rounded to only three significant figures, your calculated value of 64.28 is inaccurate. Instead, you should get 64.26 (to 4 significant figures) or 64.3 (to 3 significant figures).

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.
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