anonymous
  • anonymous
Sketch a triangle that has acute angle θ, and find the other five trigonometric ratios of θ. Tan(degree sign)= sqr of 15
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
sin(θ) = cos(θ) = csc(θ) = sec(θ) = cot(θ) =
anonymous
  • anonymous
I'll take a shot at it..... So tan is sin/cos which is y/x So x = 1 and y = sqrt15 Then do x^2 + y^2 = r^2
anonymous
  • anonymous
Does that seem right so far?

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anonymous
  • anonymous
I have no idea. I'm been trying to figure this out for a while now. Thanks for trying! If it is right kudos to you :)
anonymous
  • anonymous
I can keep going XD
anonymous
  • anonymous
So x^2 + y^2 = r^2 (r is the hypotenuse) (1)^2 + (sqrt15)^2 = r^2 1+ 15=r^2 16=r^2 sqrt16=r r=4
anonymous
  • anonymous
So now you have all the needed numbers to finish the problem. x = 1 y = sqrt15 r = 4
anonymous
  • anonymous
sin = sqrt15/4 cos = 1/4 tan = sqrt15 csc = 4/sqrt15 sec = 4 cot = 1/sqrt15
anonymous
  • anonymous
Thank you so much! But the Sin= sqrt15/4 is not the right answer.
anonymous
  • anonymous
I can't seem to figure that one out.
anonymous
  • anonymous
Is everything else right?
anonymous
  • anonymous
Yes
anonymous
  • anonymous
Is it (sqrt15)/4 or is the 4 inside the sqrt of 15?
anonymous
  • anonymous
It is (sqrt15)/4 because it is y/r.
anonymous
  • anonymous
Ok, yeah that's the right answer. Thanks again :)
anonymous
  • anonymous
Oh okay, welcome! XD
amistre64
  • amistre64
you got it?
anonymous
  • anonymous
Yeah. But I have another problem. Use the figure and the value given below to answer the following questions. Assume A = 4. (a) Express x in terms of trigonometric ratios of θ. (b) Express y in terms of trigonometric ratios of θ.
amistre64
  • amistre64
ok...gonna need the picture tho :)
amistre64
  • amistre64
or use this to let me know what I am looking at :)
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anonymous
  • anonymous
A is the Adjacent angle, y is the hypotenuse, and x is the opposite. Acute triangle.
anonymous
  • anonymous
Yes, that pic.
amistre64
  • amistre64
LIKE THIS? opps caps were on lol
amistre64
  • amistre64
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anonymous
  • anonymous
The A is on the bottom, but everything else is in the right place.
amistre64
  • amistre64
are we dealing with a right triangle?
anonymous
  • anonymous
yes
amistre64
  • amistre64
A is bottom left angle?
anonymous
  • anonymous
Just the bottom angle. I'll try to put up the pic
anonymous
  • anonymous
Here!
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amistre64
  • amistre64
ahhh...thats better, when you said A was the adjacent angle it threw me :)
anonymous
  • anonymous
Sorry :[
amistre64
  • amistre64
'sok....ill live ;)
amistre64
  • amistre64
x is equal to "sin(t)" sin(t) = x/y x = y * sin(t)
amistre64
  • amistre64
probably want my to use A for that eh....
amistre64
  • amistre64
A = 4...that helps....
anonymous
  • anonymous
I don't understand how to use it in the equation.
amistre64
  • amistre64
I think its asking for all the trig functions that use the "x" there and how we would solve for "x": x = yadayadayada stuff like that, might be overkill, but better safe than sorry :)
amistre64
  • amistre64
sin(t) = x/y ; x = y * sin(t) csc(t) = y/x ; x = y/csc(t) tan(t) = x/4 ; x = 4tan(t) cot(t) = 4/x ; x = 4/cot(t) thats all of them thatll use the x....
amistre64
  • amistre64
you know how the trig functions are defined?
anonymous
  • anonymous
Not really.
anonymous
  • anonymous
What do I do for this problem though? \
amistre64
  • amistre64
the names that are given to them arent all that important, but since they got em, might as well use em. sin is simply the ration of the opposite side from the angle divided by the hypotenuse: usually abbreviated as opp/hyp.
amistre64
  • amistre64
you just express x or y in terms of the trig functions... as far as I can tell
amistre64
  • amistre64
...simply the *ratio* of the .....
amistre64
  • amistre64
oonce you know sin and cos, the rest are just made with these two...
anonymous
  • anonymous
How would I find x and y then?
amistre64
  • amistre64
you would need more information to actually solve for x and y; you would need to know the measurement of an angle, or another side so that you could work things out. But what you have is not enough information to get any "solid" answers for the rest of it.
anonymous
  • anonymous
That's all it shows in the problem.
anonymous
  • anonymous
no angle or anything.
amistre64
  • amistre64
then it cant expect you to "solve" for any values, just want you to be able to express what x and y are in terms of the rest of the information
anonymous
  • anonymous
Here is another similar one if you can help with this one. Find the side labeled x. Assume a = 4.
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anonymous
  • anonymous
How would I express them in terms then?
amistre64
  • amistre64
this we have enough information to solve :)
amistre64
  • amistre64
well, I wrote up all the ones I could think of for the x values up there.
amistre64
  • amistre64
if a = 4 then sin(60) = x/4 sin(60) is one of those basic angles you learn in trig class and are expected to memorize. the sin(60) = sqrt(3)/2 sqrt(3) x ----- = --- 2 4 2sqrt(3) = x is as far as you can go without a calculator
anonymous
  • anonymous
Oh.. ok I get this one. But the other has me confused :P
anonymous
  • anonymous
It just says "Express x in terms of trigonometric ratios of θ."
anonymous
  • anonymous
Same for y!
amistre64
  • amistre64
the other one just says: (a) Express x in terms of trigonometric ratios of θ. (b) Express y in terms of trigonometric ratios of θ. it doesnt say "solve" for any particular value. Just use what you know about the trig functions to be able to express x or y in terms of them
amistre64
  • amistre64
like this: sin(t) = x/y ; x = y * sin(t) csc(t) = y/x ; x = y/csc(t) tan(t) = x/4 ; x = 4tan(t) cot(t) = 4/x ; x = 4/cot(t)
anonymous
  • anonymous
How would I go about doing that? I tried the X ones you put up, but they don't seem to work.
amistre64
  • amistre64
do you have anymore options :)
anonymous
  • anonymous
nope :(
amistre64
  • amistre64
then I would just skip it and move to the next question..... :)
amistre64
  • amistre64
lifes to short to worry about 1 trig problem that they dont give you enough information about right ?
anonymous
  • anonymous
Haha... Thanks so much for your help. I'm actually learning math more from you than my stupid professor lol :)
amistre64
  • amistre64
lol .... well, I do have the time.....alot of time :)
anonymous
  • anonymous
Are you in college or what?

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