Hello, topaz192!

When a passenger jet begins its descent to the airport,

it is 3900 ft from the ground. Its angle of descent is 6°. First, make a good sketch. Code:

A * - - - - - - - - - - - H
: * 6°
: *
3900 : *
: *
: 6° *
B * - - - - - - - - - - - * C

The jet is at A. .$\displaystyle AB = 3900$ feet. .The airport is at $\displaystyle C.$

$\displaystyle \angle HAC = \angle ACB = 6^o$

a) What is the plane's ground distance to the airport? We want distance $\displaystyle BC.$

In right triangle $\displaystyle ABC \!:\;\;\tan6^o = \frac{3900}{BC} \quad\Rightarrow\quad BC \:=\:\frac{3900}{\tan6^o} \:=\:37,106.02137$

Therefore: .$\displaystyle BC \;\approx\;37,106.0\text{ feet}$

b) How far must the plane fly to reach the runway? We want distance $\displaystyle AC.$

In right triangle $\displaystyle ABC\!:\;\;\sin6^o \:=\:\frac{3900}{AC} \quad\Rightarrow\quad AC\:=\:\frac{3900}{\sin6^o} \;=\;37,310.41171$

Therefore: .$\displaystyle AC \;\approx\;37,310.4\text{ feet}$