1. ## Angles of Elevation?

Ok. I missed class today so didnt quite get the lesson via math tutorial.

The question is:

When a 757 passenger jet begins its descent to the Ronald Reagan International Airport in Dc, it is 3900 ft from the ground. its angle od descent is 6 degrees.

a. what is the plane's ground distance to the airport?

b. how far must the plane fly to reach the runway?

idk how to do it from the tutorial this is what i did....

cos 6 = 3900
6 = 3921.5

is that how im suppose to folllow thru and do?

2. part (a)

$\tan(6^{\circ}) = \frac{opposite}{adjacent} = \frac{3900}{x}$

solve for x, the horizontal distance.

part (b)

$\sin(6^{\circ}) = \frac{opposite}{hypotenuse} = \frac{3900}{z}$

solve for z, the distance the airplane must fly to reach the runway.

3. 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. . $AB = 3900$ feet. .The airport is at $C.$
$\angle HAC = \angle ACB = 6^o$

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

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

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

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

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

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

4. man i have the worst problem trying to member the differences of cos, tan, and sin.

i did it all wrong cuz i did cos.

thanks yall!