I am confused if I should be using radian mode or degree mode on my calculator to find the height..
I've posted the question below. I have to find the height of the Ferris wheel from the given angle. I'm confused because
I'm solving to fill the table of values for the height, but the equation includes the 90° so I am super stumped what I should be doing.
The height above the ground of a rider on a Ferris wheel can be modeled by the sine function h(x): 25sin(x-90°) + 27,
where h(x) is the height, in metres, and x is the angle, in degrees, that the radius to the rider makes with the horizontal.
a) Complete the table for one revolution of the Ferris Wheel.
x(°) h(x) = 25sin (x - 90°) + 27 0 30 60 90 120 150 180
So.. should I be using the radian or degree mode on my calculator to solve for this?
In radian: h(0)=4.65, in degree: h(0)=2.00
I'm confused if this is basically a trick question.
Well, I do need my calculator to solve for the table of values.
And that's what I'm confused about! Let me clarify that I'm trying to solve for the height in metres, so wouldn't I use radian??
Or am I supposed to use degrees because the equation includes degrees?
Lol, I suppose so. I'm home schooled, and I have to teach myself. All the teachers for help online are not available for summer. I tried cramming in a whole unit in a week, and my last lesson for Applications of Sine Function today. So all the information is sort of muddled in my mind right now from the entire unit, since I haven't had the chance to actually go through everything slowly and take my time! I'm trying to meet college deadlines by sending in my final high school marks!
The bitter sweet feeling of procrastination! So close to the finish line, but so much stress!
If your inputs are in degrees, your calculator should be in degree mode. If your inputs are in radians, your calculator should be in radian mode. Also remember that the height h(x) is just a pure number. The unit of meters is assigned after the fact because the 25 and 27 are implicitly assumed to be 25 meters and 27 meters. This has nothing to do with the mathematical calculation; these could just as easily have been feet or miles or furlongs.
What would you do if you came upon a problem that asked you to find "5 feet minus 3 minutes"? Or "7 yards minus four pounds""? I hope you would immediately think "That can't be right!"
They can't be right because you cannot add or subtract quantities with different units. Because the problem is , the "x" must be in degrees so you can subtract "x - 90°". Now, if the problem has said then, in order to subtract " ", x would have to be in radians, just like the radians.