1. ## Trig

A lighthouse B is 8km from a point A and its bearing from A is 70 degrees E of N. A ship sails from A for 6km on a course due NE and then alters its course to 50 degrees E of N. Find from the tables its distances N and E of A when it is due North of B.

Im not sure how to do this questions. Thanks

2. Hello deltaxray
Originally Posted by deltaxray
A lighthouse B is 8km from a point A and its bearing from A is 70 degrees E of N. A ship sails from A for 6km on a course due NE and then alters its course to 50 degrees E of N. Find from the tables its distances N and E of A when it is due North of B.

Im not sure how to do this questions. Thanks
I'm not sure what 'from the tables' means. But here's how I should do it.

B is $\displaystyle 8 \sin 70^o$ km East of A and $\displaystyle 8\cos 70^o$ km North of A. (Check this out on a diagram. I'll leave you to work these values out.)

When the ship has completed the first $\displaystyle 6$ km leg, it is $\displaystyle 6\sin45^o$ km East and $\displaystyle 6\cos45^o$ km North of A. (Work these out as well.)

If it then sails a distance $\displaystyle x$ km on a bearing $\displaystyle 050^o$, it will travel a further distance $\displaystyle x\sin50^o$ km East and $\displaystyle x \cos 50^o$ km North.

If the ship is now due North of B, its distance East of A is the same as the distance that B is East of A. So:
$\displaystyle 6\sin 45^o + x\sin50^o = 8\sin 70^o$
From this we can work out the value of $\displaystyle x$ and hence how far North it is of A. (You've already worked out how far East it is.)

Can you complete the working now?

3. Hello, deltaxray!

Did you make a sketch?

A lighthouse $\displaystyle B$ is 8 km from point $\displaystyle A$
and its heading from $\displaystyle A$ is $\displaystyle N\,70^o\,E$

A ship sails from A for 6km on a course due NE to point $\displaystyle C$
and then alters its course to $\displaystyle N\,50^o\,E$ to point $\displaystyle D.$

Find its distances north and east of $\displaystyle A$ when it is due North of $\displaystyle B.$
Code:
                    :               o D
:           *   |
: 50°   *       |
:   * 40°       |
C o - - - - - - - o E
* |       3.28    |
:         *   |               o B
:     6 *     |         *     |
: 45° *       |   *           |
:   *       * |  8            |
: *   *  20°  |               |
o - - - - - - o - - - - - - - o
A    4.24     F     3.28      G
: - - - - - -7.52 - - - - - - :

We have: .$\displaystyle AB \,=\,8,\;\angle BAG \,=\,20^o$

. . . Also: .$\displaystyle AC \,=\,6,\;\angle CAF \,=\,45^o$

. . . .And: .$\displaystyle \angle DCE \,=\,40^o$

We want: .$\displaystyle \begin{Bmatrix}AG & \text{east} \\ DG & \text{north} \end{Bmatrix}$

In right triangle $\displaystyle BGA\!:\;\;\cos20^o \,=\,\frac{AG}{8} \quad\Rightarrow\quad AG \:=\:8\cos20^o \:=\:7.517590966$

. . Hence: .$\displaystyle \boxed{AG \:\approx\:7.52}$

$\displaystyle DG \:=\:DE + EG$

In right triangle $\displaystyle CFA\!:\;\sin45^o \,=\,\frac{CF}{6} \quad\Rightarrow\quad CF \,=\,6\sin45^o \,=\,4.242670687$
. . Hence: .$\displaystyle CF \:\approx\:4.24 \:=\:EG$
. . . Also: .$\displaystyle AF \:\approx\:4.24$

Then: .$\displaystyle FG \:=\:AG - AF \:=\:7.52 - 4.24 \:=\:3.28 \:=\:CE$

In right triangle $\displaystyle DEC\!:\;\tan40^o \,=\,\frac{DE}{CD} \quad\Rightarrow\quad DE \:=\:3.28\tan40^o \:=\:2.75224679$
. . So: .$\displaystyle DE\:\approx\:2.75$

Hence: .$\displaystyle DG \;=\;DE + EG \:=\:2.75 + 4.24 \quad\Rightarrow\quad\boxed{DG \:=\:6.99}$

Therefore, the ship is 6.99 km north and 7.52 km east of point $\displaystyle A.$