Hello, Juan!

I'll explain $\displaystyle N\,42^o26'\,W$

. . The $\displaystyle N$ means we start by facing *North.*

. . Then $\displaystyle 42^o26'\,W$ means we turn $\displaystyle 42^o26'$ *to the west.*

And __that__ is the direction indicated by $\displaystyle N\,42^o26'\,W$.

Similarly, $\displaystyle S30^oE$ means: face South, turn 30° to the east.

A fire is sighted due west of lookout $\displaystyle A$.

The bearing of the fire from lookout $\displaystyle B$, 8.6 miles due south of $\displaystyle A$, is $\displaystyle N\,42^o26'\,W$

How far is the fire from $\displaystyle B$ to the nearest tenth of a mile? Code:

F * - - - - - - - * A
\ |
\ |
\ |
d \ | 8.6
\ |
\ θ |
\ |
* B

The fire is at $\displaystyle F$. .Angle $\displaystyle \theta \,= \,42^o26' \,\approx \,42.4333^o$. .Let $\displaystyle d = BF.$

In right triangle $\displaystyle FAB$, we have: .$\displaystyle \cos\theta\,=\,\frac{8.6}{d}\quad\Rightarrow\quad d\,=\,\frac{8.6}{\cos\theta}$

Therefore: .$\displaystyle d \:= \:\frac{8.6}{\cos42.4333^o} \:= \:11.6521...\:\approx\:11.7$ miles.

Solve for $\displaystyle A:\;\;\sin A \:= \:\cos 3A$

Here's a back-door approach to this problem . . .

If $\displaystyle \sin A \,= \,\cos B$, then $\displaystyle A$ and $\displaystyle B$ are complementary: .$\displaystyle A + B \,=\,90^o$

Look at this right triangle. Code:

* B
c * *
* * a
* *
A * * * * * C
b

We have: .$\displaystyle \sin A = \frac{a}{c}$ .and .$\displaystyle \cos B = \frac{a}{c}$

See? . . . Angles $\displaystyle A$ and $\displaystyle B$ are from the *same right triangle.*

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Back to the problem: .$\displaystyle \sin A \:= \:\cos3A$

Since $\displaystyle A$ and $\displaystyle 3A$ are complementary: .$\displaystyle A + 3A \:= \:90^o\quad\Rightarrow\quad 4A \:=\:90^o$

Therefore: .$\displaystyle \boxed{A\:=\:22.5^o}$

Note: TPHacker's solution is __complete__.