1. ## bearings

Two lookout stations, which are 25 miles apart along the coast on a north-south line, spot an approaching yacht. One lookout station measures the direction to the yacht at N33(degrees)E, and the other station measures the direction of the yacht at S62(degrees)E. How far is the yacht from each lookout station? How far is the yacht from the coast?

(I have to show my work)

Thanks

2. Hello, chickenwing!

Two lookout stations, which are 25 miles apart along the coast on a north-south line,
spot an approaching yacht.
One lookout station measures the direction to the yacht at N33°E,
and the other station measures the direction of the yacht at S62°E.
How far is the yacht from each lookout station?
How far is the yacht from the coast?

A sketch would certainly help . . .
Code:
    A *
| *
|62°* b
|     *
|    85°* C
|      *
25 |     *
|    *
|33°* a
|  *
| *
|*
B *

The lookout stations are at A and B; the yacht is at C.
Since angle A = 62° and angle B = 33°, then angle C = 85°.

. . . . . . . . . . . . .a . . . . . . c . . . . . . . . . . 25·sin62°
Law of Sines: . ------- .= . ------- . . a .= . ----------- . . 22.16 miles
. . . . . . . . . . . sin A] . . . . sin C . . . . . . . . . .sin85°

. . . . . . . . . . . . .b . . . . . . c . . . . . . . . . . 25·sin33°
Law of Sines: . ------- .= . ------- . . b .= . ----------- . . 13.67 miles
. . . . . . . . . . . sin B . . . . sin C . . . . . . . . . .sin85°

The distance from the yacht to the coast is:
. . d .= .b·sin62° .= .13.67(sin 62°) . .12.1 miles

3. Originally Posted by chickenwing
Two lookout stations, which are 25 miles apart along the coast on a north-south line, spot an approaching yacht. One lookout station measures the direction to the yacht at N33(degrees)E, and the other station measures the direction of the yacht at S62(degrees)E. How far is the yacht from each lookout station? How far is the yacht from the coast?

(I have to show my work)

Thanks
First you need a diagram. The point B is the first point mentioned in the
problem, so the yacht has bearing 33 degrees east of north from this point.
The point A is the second point mentioned, so the yacht has a bearing of
62 degrees east of south from this point.

From the angle sum relation for a triangle we know that angle ACB is 85 degrees.

Apply the sin rule to triangle ABC to get:

sin(85)/25=sin(62)/BC=sin(33)/AC

which will allow you to find BC and AC which you require for the first part of

The final part of the question asks you to find DC. This can be done by applying
Pythagoras's theorem to triangles ADC and BDC to give:

DC^2=BC^2-BD^2
DC^2=AC^2-(25-BD)^2

which can be solved for BD and DC.

(or you can just use DC/AC=sin(62) like soroban suggests )

RonL

4. I understand! Thank you so much; Your drawing really helped me figure out where to plug in the numbers; thanks so much!!!!