highschoolmom2010
  • highschoolmom2010
find the measure of < x using trigonometric ratios
Mathematics
  • Stacey Warren - Expert brainly.com
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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.
chestercat
  • chestercat
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highschoolmom2010
  • highschoolmom2010
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highschoolmom2010
  • highschoolmom2010
@ganeshie8 IM HORRIBLE AT THESE
ganeshie8
  • ganeshie8
we'll handle this the same way... SOH-CAH-TOA go to the angle corner, and figure out given sides as opposite,adjacent and hyp

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agent0smith
  • agent0smith
Use tangent here. Tanx = opposite/adjacent. What side is opposite angle x? |dw:1376249499008:dw| \[\Large \tan x = \frac{ \text{opposite} }{\text{adjacent} }\]
highschoolmom2010
  • highschoolmom2010
opposite of
agent0smith
  • agent0smith
Hypotenuse we don't need, use tanx = opp/adj.
highschoolmom2010
  • highschoolmom2010
i know i like to list what i know :DD
agent0smith
  • agent0smith
Okay :)
agent0smith
  • agent0smith
That's a good tactic, especially for subjects like physics :)
highschoolmom2010
  • highschoolmom2010
\[\tan(x)=\frac{ 8 }{ 5 }\] \[(5)*\tan x=\frac{ 8 }{ 5 }*(5)\]
highschoolmom2010
  • highschoolmom2010
like that ?
ganeshie8
  • ganeshie8
nopes ! leave it as it is tanx = 8/5
ganeshie8
  • ganeshie8
yes ! just divide the right side, then take inverse tan both sides
highschoolmom2010
  • highschoolmom2010
i am used to having to to that i suppose
ganeshie8
  • ganeshie8
haha i bet you're !
agent0smith
  • agent0smith
No need for extra work here :P \[\tan(x)=\frac{ 8 }{ 5 }\] so \[x = \tan^{-1} \frac{ 8 }{ 5 }\]which you need a calculator for.
highschoolmom2010
  • highschoolmom2010
x\[\approx 58\]
agent0smith
  • agent0smith
Looks about right!
ganeshie8
  • ganeshie8
right right !
ganeshie8
  • ganeshie8
gw !!
highschoolmom2010
  • highschoolmom2010
so when finding angles like ^^ use tan^-1
ganeshie8
  • ganeshie8
see, you have tan x = 8/5
ganeshie8
  • ganeshie8
you want to find x, so to find x you have to isolate x
agent0smith
  • agent0smith
tan^-1 is the "opposite" of tan, so you use it to undo it. Just like a square root is the opposite of squared, a square root undoes the squard.
ganeshie8
  • ganeshie8
take tan^-1 both sides :- tan x = 8/5 tan^-1 tan x = tan^-1 8/5 x = tan^-1 8/5
highschoolmom2010
  • highschoolmom2010
oh ok that makes sense :D
ganeshie8
  • ganeshie8
left side, tan^-1 and tan cancel out
highschoolmom2010
  • highschoolmom2010
TY soo much @agent0smith && @ganeshie8
ganeshie8
  • ganeshie8
np :)
agent0smith
  • agent0smith
If you have sinx = some number, then you'd use sin^-1(some number) to undo it. Same for cosx = some number. But make sure you isolate the sinx or cosx, so it's like sinx = ... or cosx = ... ie you can't just do cos^-1 of 10cosx = 5, you'd have to divide by 10 first.

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