How do you solve this bearing trig problem?

**womatama** · May 16th 2009, 08:43 PM

A ship leaves base and sails 30 km on a bearing of 034. It is then ordered to take up position due North of the base in exactly 1 hour's time. If the ship has a speed of 20km/h, find the two possible bearings it could take.

**Grandad** · May 16th 2009, 10:34 PM

Hello womatama

Welcome to Math Help Forum!

Originally Posted by womatama

A ship leaves base and sails 30 km on a bearing of 034. It is then ordered to take up position due North of the base in exactly 1 hour's time. If the ship has a speed of 20km/h, find the two possible bearings it could take.

In the attached diagram, the ship's two possible courses are given by $BC_1$ and $BC_2$ . Then use the Sine Rule:

$\frac{\sin\theta}{30}=\frac{\sin34^o}{20}$

$\Rightarrow \sin\theta =0.8388$

There are then two possible values of $\theta: \theta_2 = 57.0^o$ and $\theta_1 = 180 - 57.0 = 123.0^o$

So the two possible courses to steer are $360 - \theta_2 = 303^o$ and $360-\theta_1 = 237^o$ .

Grandad

**womatama** · May 16th 2009, 11:43 PM

Originally Posted by Grandad

Hello womatama

Welcome to Math Help Forum!In the attached diagram, the ship's two possible courses are given by $BC_1$ and $BC_2$ . Then use the Sine Rule:

$\frac{\sin\theta}{30}=\frac{\sin34^o}{20}$

$\Rightarrow \sin\theta =0.8388$

There are then two possible values of $\theta: \theta_2 = 57.0^o$ and $\theta_1 = 180 - 57.0 = 123.0^o$

So the two possible courses to steer are $360 - \theta_2 = 303^o$ and $360-\theta_1 = 237^o$ .

Grandad

Oh thanks alot !! i drew the thing but didnt know what to do lol. Thanks.
One question: How come you took 57 away from 180 to get the other value? i dont understand that

**Grandad** · May 17th 2009, 12:17 AM

Hello womatama

Originally Posted by womatama

Oh thanks alot !! i drew the thing but didnt know what to do lol. Thanks.
One question: How come you took 57 away from 180 to get the other value? i dont understand that

I don't know how much you know about sines and cosines, but amng the rules are, for any value of $\theta$ :

$\sin\theta = \sin(180-\theta)$

$\cos\theta = \cos(180-\theta)$

which tells you how to find the sine and cosine of obtuse angles. So, for example, $\sin57^o = \sin(180-57)^o=\sin123^o$ (and $\cos 123^o = -\cos57^o$ ). Check these out on your calculator if you like.

Grandad

**womatama** · May 17th 2009, 01:26 AM

Oh i get it now
i looked at the sine graph and cosine graph and then i understood it. THX

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