A skier descends a mountain at an angle of 35.0º to the horizontal. If the mountain is 235 m long, what are the horizontal and vertical components of the skier's displacement?

3 answers

I hope you mean the ski track along the mountain is 235 m long.

The mountain can be represented by a right triangle with the two legs horizontal and vertical, and where the angle of elevation is 35°.
The ski track is the hypotenuse (C), and if the horizontal leg (B) and the vertical leg (A), then
A=C(sin(35°))
B=C(cos(35°))
You can use your calculator or a trigonometric table to find sin(35°) and cos(35°).
Excuuuuuse me! How long ago did you take trig? No one uses tables any more.

However, the trig book I used was titled "Trigonometry with Tables."

I remember our library had huge volumes from the National Bureau of Standards with 12-place tables for trig, log, and exponential functions (in case your slide rule wasn't good enough).
Haha! At least someone noticed that!

Yes, in my days, slide-rules were not accurate enough, and survey calculations were done using 7-figure (log and trig) tables which were packed in a volume (I still have one!) thicker than today's calculus textbook. That taught me how to interpolate rapidly in the head.

In parallel with slide-rules, 4-function mechanical calculators were also used:
http://upload.wikimedia.org/wikipedia/commons/thumb/5/57/Odhner_made_before_1900.jpg/220px-Odhner_made_before_1900.jpg
I still have a portable version of the above and on which I have devised an algorithm to find square-roots accurate to the capacity of the machine (not by iterations)!

I have also used addiators similar to this:
http://upload.wikimedia.org/wikipedia/commons/thumb/5/57/Odhner_made_before_1900.jpg/220px-Odhner_made_before_1900.jpg

To make a long story short, I included the option of trig tables because not all users of Jiskha come from the US, and in many other countries, calculators may not be widely available.