Trig help needed!

Original post by SherlockHolmes

Which length are you struggling to work out? It looks like you got all the information you need to solve the problem.

Sorry! I meant the angle

Reply 3

The length is correct.

To find the angle

\theta_a

use the sine rule. Since you know a side and an opposite already (the side and angle you just worked out).

But then since this is the bearing you'll need to take away the angle

\theta_a

and that angle next to

\theta_a

and take both away from

360^o

(edited 11 years ago)

Reply 4

SherlockHolmes

Using the angle 110 and length of side 16.5km, you can use the sine rule to work out either of the other two angles.

Reply 5

the bear

You should use the cosine rule with:

a = ?
b = 12
c = 8
A = 30 + ( 180 - 100 )

to find the bearing find angle B and add 30^o

Reply 6

Original post by Secret.

You need to use the sine rule for the angle

Not sure how you are getting 46

The answer seems to be 057

Reply 7

Original post by SherlockHolmes

Using the angle 110 and length of side 16.5km, you can use the sine rule to work out either of the other two angles.

Original post by the bear

You should use the cosine rule with:

a = ?
b = 12
c = 8
A = 30 + ( 180 - 100 )

to find the bearing find angle B and add 30^o

Original post by TenOfThem

You need to use the sine rule for the angle

Not sure how you are getting 46

The answer seems to be 057

I'm getting the wrong answer doing:

Sin A / 12 = Sin (110) / 16.5
and rearranging gives an answer of 43.11..., which values have I gotten wrong here?

Reply 8

Original post by Secret.

I'm getting the wrong answer doing:

Sin A / 12 = Sin (110) / 16.5
and rearranging gives an answer of 43.11..., which values have I gotten wrong here?

That is correct but A is not the angle that you want

Reply 9

Original post by Secret.

I'm getting the wrong answer doing:

Sin A / 12 = Sin (110) / 16.5
and rearranging gives an answer of 43.11..., which values have I gotten wrong here?

TenofThem did

\frac {sin(b)}{8} = \frac {sin(110)}{16.5}

You've used the sine rule to find

\theta_a

straight away, which is just valid.

Now that you've got angle

\theta_a

, do the standard procedure to turn that into a bearing.

Reply 10

Original post by TenOfThem

That is correct but A is not the angle that you want

Ohh right, I got the answer now thanks!

Also, why does it give wrong answer if I tried to calculate A directly?

Reply 11

Original post by Secret.

Ohh right, I got the answer now thanks!

Also, why does it give wrong answer if I tried to calculate A directly?

It doesn't, you've just got a different angle.

Reply 12

Original post by Secret.

Ohh right, I got the answer now thanks!

Also, why does it give wrong answer if I tried to calculate A directly?

Not sure what you mean?

Reply 13

Original post by 0x2a

TenofThem did

\frac {sin(b)}{8} = \frac {sin(110)}{16.5}

You've used the sine rule to find

\theta_a

straight away, which is just valid.

Now that you've got angle

\theta_a

, do the standard procedure to turn that into a bearing.

I'm sorry but how would I do that? From 43.11... :O?

(edited 11 years ago)

Reply 14

Original post by Secret.

I'm sorry but how would I do that? From 043.11... :O?

The angle on top of

\theta_a

80^o

due to interior angles adding up to 180 degrees.

Now take 80 degrees and angle

\theta_a

away from 180, and you get 57 degrees. Then add 180 to this and you've got your bearing.

Reply 15

Original post by Secret.

I'm sorry but how would I do that? From 43.11... :O?

Do you know what angle you are actually trying to find?

Reply 16

Original post by TenOfThem

Do you know what angle you are actually trying to find?

b, I think? I was trying to work out a before that's why I was getting a different answer, but now I'm confused even more :P

Original post by 0x2a

The angle on top of

\theta_a

80^o

due to interior angles adding up to 180 degrees.

Now take 80 degrees and angle

\theta_a

away from 180, and you get 57 degrees. Then add 180 to this and you've got your bearing.

Sorry, slightly confused but isn't the angle on top of theta a, 110? or do you mean If I extend the north line down and make that in to a separate triangle?

Reply 17

Original post by Secret.

b, I think?

You are trying to find 30+b

That is the bearing of the 16.5 line

Reply 18

Original post by Secret.

b, I think?

Sorry, slightly confused but isn't the angle on top of theta a, 110? or do you mean If I extend the north line down and make that in to a separate triangle?

Yup, that's what I was getting at :tongue:

Just getting angle 30 + B and adding it on to 180 degrees (alternate angles) is a much faster way though. But it's good to have knowledge of many ways to solve a particular problem.

Reply 19