# Thread: Law of Sines....I think

1. ## Law of Sines....I think

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?

2. Hello, suzdisp!

Did you make a sketch?

Two lookout stations, which are 25 miles apart along a north-south costline,
spot an approaching yacht.
One lookout station finds the direction to the yacht at N33°E,
and the other station finds the direction of the yacht at S62°E.

(1) How far is the yacht from each lookout station?

(2) How far is the yacht from the coast?
Code:
  -  B*
:   |  *
:   | 62° *   a
:   |        *
:   |           *
:   |              *
:  D* - - - - - - 85° *C
:   |               *
25   |             *
:   |           *
:   |         *
:   |       * b
:   |     *
:   |33°*
:   | *
-  A*

Since $\displaystyle \angle A = 33^o,\:\angle B = 62^o$, then: .$\displaystyle \angle C \:=\:180^o - 33^o - 62^o \:=\:85^o$

Law of Sines: .$\displaystyle \frac{a}{\sin A} \:=\:\frac{c}{\sin C} \quad\Rightarrow\quad a \:=\:\frac{c\sin A}{\sin C}$

. . $\displaystyle a \:=\:\frac{25\sin33^o}{\sin86^o} \;=\;13.66798666 \quad\Rightarrow\quad\boxed{ a \;\approx\;13.7\text{ miles}}$ (1)

Law of Sines: .$\displaystyle \frac{b}{\sin B} \:=\:\frac{c}{\sin C} \quad\Rightarrow\quad b \:=\:\frac{c\sin B}{\sin C}$

. . $\displaystyle b \:=\:\frac{25\sin62^o}{\sin85^o} \:=\:22.15800773 \quad\Rightarrow\quad\boxed{ b \;\approx\;22.2\text{ miles}}$ (1)

In right triangle $\displaystyle CDA\!:\;\;\sin33^o \:=\:\frac{CD}{b}$

Therefore: .$\displaystyle CD \:=\:22.2\sin33^o \:=\:12.09068658\quad\Rightarrow\quad\boxed{ CD \:\approx\:12.1\text{ miles}}$ (2)