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)**