help
A ladder rest against a wall forming 40’ angle with floor. Use the function y=8 sec theta to find the length of the ladder to the nearest foot.
I'm guessing that y is the hypotenuse of the triangle. 8 Feet is opposite of theta but if we are using secant it should be adjacent.|dw:1401586772092:dw|
Is there any other information with this problem?
no
Okay well do we agree with my drawing of the problem?
yes
I'm retyping that to make it clearer **************************************************** Well the COSECANT of 40° = hyp / opp COSECANT of 40° = hyp / 8 hypotenuse = csc (40°) * 8
hypotenuse = 1.5557 * 8 hypotenuse (or ladder height) = 12.4456
where you get the 1.5557
It is the cosecant of 40°
ok
I see now
so the nearest foot is 12 ft
The answer is okay with you. As I said I changed the problem. If I worked with secant instead of cosecant, the answer would not be the same. Secant (40°) = 1.3054 So answer would be 1.3054 * 8
Oh geez - I forgot to round it - yes it rounds to 12 (I guess I felt it was more important to get the problem worded correctly.)
10ft
Why did you round it to 10 feet? I spent all that time explaining how the problem was worded INCORECTLY.
1.3054*8=10.44
Yes but I added that at the end to say that is the incorrect answer. By the way how familiar are you with trig?
Okay - based on the INCORRECT way the problem is stated then the INCORRECT answer is 10. Maybe everyone else in your class (including the instructor) will say 10.
but the correct is 12 ft
I am just guessing that the problem is worded INCORRECTLY. I say the problem should be using COSECANT and NOT secant. The answer should be 12. Show it to your instructor if he says it is wrong. Suppose (as an example) that a problem says Jack Smith (20 years old) has a SON who is 40 years old , etc Wouldn't you say there's an error in there?
yes and just to return to the Jack Smith problem if we answered that Jack Smith (the 20 year old father) has a son who is 40 years old then the son is -20 years (that's MINUS 20 years) younger than the father. Okay let's head on to the new problem. :-)
If a pebble is dropped into a pond in the shape of an ellipse at the location of one focus, the waves will converge at the location of the other focus. If the pond has a major axis of 20 feet and a minor axis of 12 feet, how far apart are the foci?
Wow - I have a web page that can calculate ellipses. The foci distance = 16 feet http://1728.org/ellipse.htm I can show the math (which you will probably will need right?)
not sure how to input the info
Okay where it says "Which two items of data do you want to input?" Click on the box "Major Axis and Minor Axis" Then you'll see the 2 boxes that need to be filled in. If you need more help, let me know.
but how do you do it manually?
can you use it to find the foci of a hyperbola with an equation?
No - I do not have a hyperbola calculator. You want the actual ellipse mathematics right? Are you familiar with the Pythagorean Theorem?
a^2+b^2=c^2
yes !!!! that is the pyth theorem
yep
okay - geez I have phone call I will return here - about 30 minutes promise
ok
Link to wikipedia http://en.wikipedia.org/wiki/Ellipse Also, look at the diagram at the upper left on this page: http://1728.org/ellipse.htm Line Cf1 = half the major axis Line CO = half the minor axis Major Axis = 20 feet Minor Axis = 12 feet Line Cf1 = 10 feet Line CO = 6 feet From Pythagorean Theorem: Line (f1O)² = (Cf1)² -(CO)² Line (f1O)² = 10² -6² Line (f1O)² = 100 -36 Line (f1O)² = 64 Line f1O = 8 Foci distance = Line f1f2 = 2 * 8 = 16 There you go - and yes it is just that "simple" LOL good luck
Thanks
u r welcome
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