# vectors-bearings-trig

• Jan 24th 2010, 03:48 AM
protractor
vectors-bearings-trig
hello,
I haven't found out how to complete my profile yet, so- I am a 54 year old woman, with 2 grown up children, I study Maths at home, mainly from 2nd hand Maths books I've bought in charity shops, although sometimes I have to buy a new one. I was average at Maths at school, just about got an O level, but have become much much better as an adult, and I'm absolutely sure that by dint of persistent studying my brain has actually physically changed, I'm so much better now!

But now I'm stuck again . It's a question from an AS Maths text book. I've tried to draw a diagram of it, but I keep getting confused and going round in circles and I'm not sure what to put where, it's worse than being stuck out in the Pacific for days without food or water, not knowing which way to turn, and I'm becoming hysterical over it(Wink)
Here it is:

Find the magnitude and direction of the resultant of the two given vectors:

A displacement of magnitude 3.5 km on a bearing 50 degrees, and a displacement of magnitude 5.4 km on a bearing of 128 degrees.

In the text book, the ANSWER is given as: 7.02 km and 98.9 degrees.
• Jan 24th 2010, 04:54 AM
Hello protractor

Welcome to Math Help Forum!
Quote:

Originally Posted by protractor
hello,
I haven't found out how to complete my profile yet, so- I am a 54 year old woman, with 2 grown up children, I study Maths at home, mainly from 2nd hand Maths books I've bought in charity shops, although sometimes I have to buy a new one. I was average at Maths at school, just about got an O level, but have become much much better as an adult, and I'm absolutely sure that by dint of persistent studying my brain has actually physically changed, I'm so much better now!

But now I'm stuck again . It's a question from an AS Maths text book. I've tried to draw a diagram of it, but I keep getting confused and going round in circles and I'm not sure what to put where, it's worse than being stuck out in the Pacific for days without food or water, not knowing which way to turn, and I'm becoming hysterical over it(Wink)
Here it is:

Find the magnitude and direction of the resultant of the two given vectors:

A displacement of magnitude 3.5 km on a bearing 50 degrees, and a displacement of magnitude 5.4 km on a bearing of 128 degrees.

In the text book, the ANSWER is given as: 7.02 km and 98.9 degrees.

Take a look at the diagram I've attached. You'll see that each angle is measured clockwise from the North. Can you see that this will give:
$\angle ABC = 360 - (130 + 128) = 102^o$
So now you use the Cosine Rule on the triangle to get:
$AC^2 = 3.5^2 + 5.4^2 - 2\times3.5\times5.4\cos102^o$
$AC = 7.02$ km (Can you work that out?)
Then use the Sine Rule to work out $\angle BAC$, and add $50^o$ to get the bearing of $AC=098.8^o$.

Can you complete it now?

• Jan 24th 2010, 04:56 AM
Hi,

if you draw a sketch of the situation,
with the 3.5 length arrow at 50 degrees and the 5.4 length arrow at 128 degrees, then you see that the longer one makes an angle of 52 degrees with the negative x-axis.
Draw a parallelogram such that the resultant vector starts at the origin
and touches the top point of the parallelogram.

The obtuse angle at the right side of the parallelogram is 50+52=102 degrees.

Now you can use the Law of Cosines (Cosine Rule) to calculate the
length of the resultant vector.

$L=\sqrt{(3.5)^2+(5.4)^2-(3.5)95.4)Cos102^o}=7.019$

Now that you have found it's length, you can calculate the acute angle
between it and either original vector using the Law of Sines (Sine Rule).

• Jan 24th 2010, 04:58 AM
Sudharaka
Dear protractor,

I have drawn the vector diagram for you. Try to solve the problem from here. If you have further suggesions please don't hesitate to reply me.
• Jan 24th 2010, 11:00 AM
protractor
merci beaucoup
Quote:

Hello protractor

Welcome to Math Help Forum!Take a look at the diagram I've attached. You'll see that each angle is measured clockwise from the North. Can you see that this will give:
$\angle ABC = 360 - (130 + 128) = 102^o$
So now you use the Cosine Rule on the triangle to get:
$AC^2 = 3.5^2 + 5.4^2 - 2\times3.5\times5.4\cos102^o$
$AC = 7.02$ km (Can you work that out?)
Then use the Sine Rule to work out $\angle BAC$, and add $50^o$ to get the bearing of $AC=098.8^o$.

Can you complete it now?

Thank you so much for your explanation, with the diagram Grandad, that's brilliant, and I completely understand it now. Yesterday, when I was struggling to do this question, I came across the old "interior angles in parallel lines add up to 180 degrees" etc, which reminded me of when I was 12 at school, and also enabled me to understand your explanation. When I get a clear explanation like that, it reminds me of why I enjoy Maths so much, instead of it causing me to go beserk, as it did yesterday.

I'm now about to examine Archie Meade's and Sudharaka's explanations, in which I spot the word "parallelogram" which rings a bell - I think I remember seeing something in a GCSE book about how parallelograms can be used, as well as triangles, for vectors, .
• Jan 24th 2010, 12:40 PM
protractor
interesting pathways and turnings
Quote:

Originally Posted by Sudharaka
Dear protractor,

I have drawn the vector diagram for you. Try to solve the problem from here. If you have further suggesions please don't hesitate to reply me.

There, I've just done it the parallelogram way! Thanks Archie Meade and Sudharaka for your interesting alternative method of finding the answer. I found your diagram very helpful Sudhharaka, it helped me to picture the whole thing, and I copied it and wrote in angles and line measurements etc.

Next, I intend to practise it both the Grandad and the Archie Meade/Sudharaka way, until hopefully I've completely mastered it. Thank you so much.