Hello, AC!
The second problem took some manuvering . . .
2. A 300ft broadcast antenna stands at the tip of a hill
whose sides are inclined at 18.6° to the horizontal.
How far down the the hill will a 250 ft. support cable extend
if it is attached halfway up the antenna? Code:
o B

 150

o C
* 
250 *  150
* 
* o D
F * * :
o x :
* :
* 18.6° :
A o               o E
$\displaystyle \Delta DEA$ is the hill. .$\displaystyle \angle A = 18.6^o$
$\displaystyle BD$ is the antenna: .$\displaystyle BC = 150,\;CD = 150$
$\displaystyle CF$ is the cable: .$\displaystyle CF = 250$
We want: .$\displaystyle x = FD$
In right triangle $\displaystyle DEA\!:\;\angle A = 18.6^o \quad\Rightarrow\quad \angle ADE = 71.4^o \quad\Rightarrow\quad \angle CDF = 108.6^o $
In $\displaystyle \Delta CDF$, use the Law of Sines: .$\displaystyle \frac{\sin(\angle CFD)}{150} \:=\:\frac{\sin108.6^o}{250} $
. . $\displaystyle \sin(\angle CFD) \:=\:0.568661046 \quad\Rightarrow\quad \angle CFD \:\approx\:34.6^o$
Then: .$\displaystyle \angle C \:=\:180^o  108.6^o  34.6^o \;=\;36.7^o$
In $\displaystyle \Delta CDF$, we have: .$\displaystyle \frac{x}{\sin36.7^o} \:=\:\frac{250}{\sin108.6^o} $
. . Therefore: .$\displaystyle x \;=\;157.6400787 \;\approx\;157.6\text{ ft} $