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.
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.
Hello womatama
Welcome to Math Help Forum!In the attached diagram, the ship's two possible courses are given by $\displaystyle BC_1$ and $\displaystyle BC_2$. Then use the Sine Rule:
$\displaystyle \frac{\sin\theta}{30}=\frac{\sin34^o}{20}$
$\displaystyle \Rightarrow \sin\theta =0.8388$
There are then two possible values of $\displaystyle \theta: \theta_2 = 57.0^o$ and $\displaystyle \theta_1 = 180 - 57.0 = 123.0^o$
So the two possible courses to steer are $\displaystyle 360 - \theta_2 = 303^o$ and $\displaystyle 360-\theta_1 = 237^o$.
Grandad
Hello womatamaI don't know how much you know about sines and cosines, but amng the rules are, for any value of $\displaystyle \theta$:
- $\displaystyle \sin\theta = \sin(180-\theta)$
- $\displaystyle \cos\theta = \cos(180-\theta)$
which tells you how to find the sine and cosine of obtuse angles. So, for example, $\displaystyle \sin57^o = \sin(180-57)^o=\sin123^o$ (and $\displaystyle \cos 123^o = -\cos57^o$). Check these out on your calculator if you like.
Grandad