Trig

**deltaxray** · February 23rd 2010, 02:28 AM

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

**Grandad** · February 23rd 2010, 03:04 AM

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 $8 \sin 70^o$ km East of A and $8\cos 70^o$ 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 $6$ km leg, it is $6\sin45^o$ km East and $6\cos45^o$ km North of A. (Work these out as well.)

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

$6\sin 45^o + x\sin50^o = 8\sin 70^o$

From this we can work out the value of $x$ 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?

Grandad

**Soroban** · February 23rd 2010, 07:06 AM

Hello, deltaxray!

Did you make a sketch?

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

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

Find its distances north and east of $A$ when it is due North of $B.$

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

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

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

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

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

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

$DG \:=\:DE + EG$

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

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

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

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

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

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