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find the measure of < x using trigonometric ratios

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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

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Other answers:

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