anonymous
  • anonymous
a right triangle has acute angle theta and cos theta = 2/5. what are the other 5 trigonometric ratios of theta?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rosedewittbukater
  • rosedewittbukater
|dw:1371061540347:dw| The ratio for cosine is adjacent/hypotenuse. You can remember the ratios by thinking of SOH CAH TOA. (sin = opp/hyp, cos = adj/hyp, tan = opp/adj). Since the cosine is adjacent over hypotenuse, and the cos theta in the triangle is 2/5, you now know what the adjacent and hypotenuse are. Then use the pythagorean theorem to find the last side. a^2 + b^2 = c^2. A is the adjacent side, and c is the hypotenuse. See if you can find b.
anonymous
  • anonymous
b = 5.4?
anonymous
  • anonymous
You can find the rest of the things
1 Attachment

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

rosedewittbukater
  • rosedewittbukater
Close. Did you plug in for a and c? \[2^2+b^2=5^2\]\[4+b^2=25\]\[b^2=21\]\[\sqrt{21}=4.58\]
anonymous
  • anonymous
I have no idea what i did lol I've been doing math constantly for 5 days straight, which is why I get confused and lost so easily. Sorry, keep going please.
rosedewittbukater
  • rosedewittbukater
Lol I know what you mean. It's ok. So now that you have b=4.58, that is the opposite of the theta angle. The 6 ratios are cosine, sine, tangent, secant, cosecant, and cotangent. You already have cosine. Remember SOH CAH TOA for sine and tangent. Secant(sec) is the reciprocal of cosine, so instead of adj/hyp, it is hyp/adjacent. Cosecant(csc) is the reciprocal of sine, so it is hyp/opp. Cotangent is the reciprocal of tangent, so adj/opp.
anonymous
  • anonymous
So I just plug in all my values and write them as fractions?
rosedewittbukater
  • rosedewittbukater
Yes. Remember that the sides are to the theta angle. So the hypotenuse is the long side(5), the adjacent side is 2, and the opposite is 4.58.
rosedewittbukater
  • rosedewittbukater
Sometimes it's easier to draw it out so you don't get mixed up. That's what I do. |dw:1371062368134:dw|
anonymous
  • anonymous
Thank you! While I type all that out, can you help me with this one? http://puu.sh/3eb1A.png
rosedewittbukater
  • rosedewittbukater
Your welcome! Typing out math sucks.. Idk why teachers make us do that :/
anonymous
  • anonymous
i don't get it either. /
anonymous
  • anonymous
cos40º = adjacent side / hypotenuse cos40º = 8000/d d = 8000/cos40º The question give the value of cos40º :]
rosedewittbukater
  • rosedewittbukater
Ok so since you already have 2 of the angles, just subtract them both from 180 to find the last angle. 180-90-40=50. You'll probably want to do the law of sines on this one. Basically the law of sines is a ratio of the sine of the angle over its side is equal to the same ratio of another side and angle... \[\frac{ \sin 50^{o} }{ 8000 }=\frac{ \sin 90^{o} }{ d }\] After that you cross multiply. So \[\sin50^{o}d=\sin90^{o}(8000)\]Use a graphing calculator to find simplify. If you do sin 50 degrees, make sure to put the degree sine on the calculator. Press 2nd, angle, and 1 to put the degree sign. Sin90deg is 1 because of the unit circle. So simplify and solve for d.
rosedewittbukater
  • rosedewittbukater
Zair's way works too. It's probably simpler lol but I do it this way out of habit because I always get mixed up with the law of cosine.
anonymous
  • anonymous
D = 6160, but that doesnt make sense because D is longer than the side with 8000 ft.
rosedewittbukater
  • rosedewittbukater
How ddi you get 6160?
anonymous
  • anonymous
cos40º = 0.77 Do 8000/0.77 = 10389,61
rosedewittbukater
  • rosedewittbukater
I think zair's way is easier. Sorry for confusing you.
anonymous
  • anonymous
i did it backwards. duh. omg. lol thank you
rosedewittbukater
  • rosedewittbukater
No problem
anonymous
  • anonymous
LAST ONE: http://puu.sh/3ebAa.png
anonymous
  • anonymous
angle a = 51
rosedewittbukater
  • rosedewittbukater
How did you get angle a? 180-90-31=59.
anonymous
  • anonymous
oops, read it from my paper wrong. lol i'm all over the place today
rosedewittbukater
  • rosedewittbukater
|dw:1371063372783:dw|
anonymous
  • anonymous
It's more easy to only the angle 31º. sen31º = b / 14 b = 14*sen31º cos31º = a/14 a = 14*cos31º
anonymous
  • anonymous
only use*
anonymous
  • anonymous
b = 7.2 a = 12
anonymous
  • anonymous
yep!
rosedewittbukater
  • rosedewittbukater
yes that is correct

Looking for something else?

Not the answer you are looking for? Search for more explanations.