# Thread: How do you solve this bearing trig problem?

1. ## How do you solve this bearing trig problem?

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.

2. 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 $360-\theta_1 = 237^o$.

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 $360-\theta_1 = 237^o$.

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

4. 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.