highschoolmom2010
find the measure of < x using trigonometric ratios
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highschoolmom2010
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@ganeshie8 IM HORRIBLE AT THESE
ganeshie8
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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
agent0smith
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Use tangent here. Tanx = opposite/adjacent. What side is opposite angle x? |dw:1376249499008:dw|
\[\Large \tan x = \frac{ \text{opposite} }{\text{adjacent} }\]
highschoolmom2010
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opposite of <x 8
adjacent to <x 5
hypotenuse unknown
agent0smith
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Hypotenuse we don't need, use tanx = opp/adj.
highschoolmom2010
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i know i like to list what i know :DD
agent0smith
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Okay :)
agent0smith
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That's a good tactic, especially for subjects like physics :)
highschoolmom2010
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\[\tan(x)=\frac{ 8 }{ 5 }\]
\[(5)*\tan x=\frac{ 8 }{ 5 }*(5)\]
highschoolmom2010
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like that ?
ganeshie8
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nopes ! leave it as it is tanx = 8/5
ganeshie8
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yes ! just divide the right side, then take inverse tan both sides
highschoolmom2010
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i am used to having to to that i suppose
ganeshie8
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haha i bet you're !
agent0smith
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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
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x\[\approx 58\]
agent0smith
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Looks about right!
ganeshie8
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right right !
ganeshie8
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gw !!
highschoolmom2010
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so when finding angles like ^^ use tan^-1
ganeshie8
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see, you have
tan x = 8/5
ganeshie8
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you want to find x, so to find x you have to isolate x
agent0smith
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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
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take tan^-1 both sides :-
tan x = 8/5
tan^-1 tan x = tan^-1 8/5
x = tan^-1 8/5
highschoolmom2010
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oh ok that makes sense :D
ganeshie8
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left side, tan^-1 and tan cancel out
highschoolmom2010
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TY soo much @agent0smith && @ganeshie8
ganeshie8
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np :)
agent0smith
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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.