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?
To find the horizontal and vertical components of the skier's displacement, you can use trigonometry. The horizontal component is the side adjacent to the angle, while the vertical component is the side opposite to the angle.
First, let's find the horizontal component of the displacement. We can use the cosine function, which is defined as the adjacent side divided by the hypotenuse.
cos(angle) = adjacent / hypotenuse
Plugging in the values we have:
cos(35.0º) = horizontal component / 235 m
Now solve for the horizontal component:
horizontal component = cos(35.0º) * 235 m
Calculate the horizontal component:
horizontal component = 0.819 * 235 m
horizontal component ≈ 192.51 m
Next, let's find the vertical component of the displacement. We can use the sine function, which is defined as the opposite side divided by the hypotenuse.
sin(angle) = opposite / hypotenuse
Plugging in the values we have:
sin(35.0º) = vertical component / 235 m
Now solve for the vertical component:
vertical component = sin(35.0º) * 235 m
Calculate the vertical component:
vertical component = 0.574 * 235 m
vertical component ≈ 134.89 m
Therefore, the horizontal component of the skier's displacement is approximately 192.51 m, and the vertical component is approximately 134.89 m.
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.
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°).