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Idk what to do, I made a triangle for the problem, but when it came to angle of elevation im clueless...help?
The rim of a basketball hoop is 10ft above the ground. The free-throw line is 15 feet from the basket rim. If the eyes are of a basketball player are 6 feet above the ground, what is the angle of elevation of the players line of sight when shooting a free throw to the rim of a basket?
http://i20.photobucket.com/albums/b2..._188/math2.jpg
refer to the diagram below. you want to find the angle xQuote:
Originally Posted by topaz192
think "trig ratio." you have the opposite side and the adjacent side to the angle x
can you continue?
So all I do is TAN = 10 / 15? put that in the calculator and thats it?
you didn't look at my diagram, did you? there is no 10 anymore. please view the diagram i postedQuote:
Originally Posted by topaz192
Sorry, I assumed it was still 10.
Why is it now 4?
TAN = 4 / 15
first off, trig functions do not stand alone. you cannot just have tan you must have tan(something). in this case, the something is xQuote:
Originally Posted by topaz192
note the modification i made to your original diagram. the broken red line is the base of the triangle we are working with. it is 6 feet off the ground, so we only go 4 more feet to hit the rim. this forms the height of our right-triangle
so you have tan(x) = 4/15 and now we continue
Oh okay I see what you're saying. (Giggle)
So in the end it'll be about 14.9 degrees.
yesQuote:
Originally Posted by topaz192
Thank you for your time! Now I understand how to solve problems like this! Yay! (Wink)