Law of Sines....I think

Hello, suzdisp!

Did you make a sketch?

Quote:

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

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

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

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

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

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

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