
Trigonometry problems...
Greetings Everyone.
I have a few problems that I dont understand and was wondering if anyone here could help break it down for me so its easier for me to understand and solve. My teacher told me that I needed some knowledge of precalculus to do this (Law of Cosines and stuff). I'm taking Precalculus right now so I wouldn't know. (Itwasntme)
Here are the questions
1. The Gondola ski lift at keystone, colorado is 2830 m long. On average, the ski lift rises 14.6 degrees above the horizontal. How high is the top of the ski lift relative to the base?
2. A highway is to be built between two towns, one of which lies 35.0km south and 72.0 km west of the other. What is the shortest length of highway that can be built between the two towns, and at what angle would this highway be directed with respect to due west?
3. A hill that has 12.0% grade is one that rises 12.0m vertically for every 100.0m of distance in the horizontal direction. At what angle is such a hill inclined above the horizontal?
Thanks in advance.

No law of cosines or any "precalculus" stuff ... this is just basic right triangle trig. Did you work problems like this back in Geometry class?
Note that it's important you draw a sketch of each situation to visualize the sides of the right triangle.
1. The Gondola ski lift at keystone, colorado is 2830 m long. On average, the ski lift rises 14.6 degrees above the horizontal. How high is the top of the ski lift relative to the base?
hypotenuse = 2830 m
you want the height which would be the side opposite the 14.6 degree angle. use the sine ratio ...
$\displaystyle \sin{\theta} = \frac{opposite}{hypotenuse}$
substitute your known values and solve.
2. A highway is to be built between two towns, one of which lies 35.0km south and 72.0 km west of the other. What is the shortest length of highway that can be built between the two towns, and at what angle would this highway be directed with respect to due west?
find the length using Pythagoras.
for the angle, 35 km is the opposite side and 72 km is the adjacent side.
use the tangent ratio ...
$\displaystyle \tan{\theta} = \frac{opposite}{adjacent}$
to get $\displaystyle \theta$ , use the inverse tangent function ...
$\displaystyle \theta = \tan^{1}\left(\frac{opposite}{adjacent}\right)$
3. A hill that has 12.0% grade is one that rises 12.0m vertically for every 100.0m of distance in the horizontal direction. At what angle is such a hill inclined above the horizontal?
same drill as #2.

In Geometry I don't really remeber doing that stuff. I checked my Geometry notebook and they have no notes on it. (Thinking) I think we should have learned it but my teacher had to go on maternity leave so we were stuck with a sub for the last 2 months of school.

For the first one I got 713m high. Not really sure tho.
The second one I got 26 degrees. Is that right?
I got 6.8 degrees for the third question.
Can anyone please check over my answers? (Wink)

those solutions are fine.

Yay (Rofl)
Thank you for checking them (Happy)