anonymous
  • anonymous
Explain SOH-CAH-TOA sin=______ cos=______ tan=_______
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@peachpi
triciaal
  • triciaal
|dw:1440640439967:dw|
triciaal
  • triciaal
|dw:1440640548365:dw|

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anonymous
  • anonymous
O thanks :)
triciaal
  • triciaal
SOH -CAH-TOA is a "code" used to remember the trig identities this is with a right triangle the hypotenuse is always opposite the right angle and is the longest side of the triangle
anonymous
  • anonymous
If i wanted to know sin cos and tan but in terms of not using opp hyp and adj. what would they look like?
anonymous
  • anonymous
@triciaal
anonymous
  • anonymous
they would be points on the unit circle where the input was the angle, and the output for cosine is the first coordinate and for sine the second coordinate of that point
anonymous
  • anonymous
anonymous
  • anonymous
i kind of dont understand
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
if you are using the mnemonic device soh cah toa , save what i sent for later it will be a big help believe me, but you are not ready for it yet
anonymous
  • anonymous
ok
anonymous
  • anonymous
i just dont know what to put in the blanks in number 5
anonymous
  • anonymous
oh i think they want you to say\[\sin(\theta)=\frac{y}{r},\cos(\theta)=\frac{x}{r},\tan(\theta)=\frac{y}{x}\]
anonymous
  • anonymous
whoever wrote this made a very serious error however
anonymous
  • anonymous
what was the error? It was my teacher btw lol
anonymous
  • anonymous
sine is a function, it is spelled "sine" if you want to write it in function notation, it needs to be \(\sin(\theta)\) or \(\sin(\xi)\) or \(\sin(\clubsuit)\) or something with a variable in the parentheses not "sin = " which makes no sense at all
anonymous
  • anonymous
you can tell him/her satellite said they were abusing notation which will lead to massive confusion for you down the road
anonymous
  • anonymous
Oh, ill be sure to tell her. What would i label the side lying on the x axix and the vertical side??
anonymous
  • anonymous
x and y i guess
anonymous
  • anonymous
|dw:1440642321127:dw|
anonymous
  • anonymous
|dw:1440642400091:dw|
anonymous
  • anonymous
|dw:1440642460037:dw|
anonymous
  • anonymous
is that it?
anonymous
  • anonymous
yeah looks good
anonymous
  • anonymous
btw in the picture you have a \(\theta\) as the angle, so it should be \(\sin(\theta)\) etc also note that \(r=\sqrt{x^2+y^2}\) by pythagoras
anonymous
  • anonymous
all right. Thanks. can you guide me with this?
triciaal
  • triciaal
#6 use pythagoren theorem @satellite73 has it above
anonymous
  • anonymous
yeah we just anticipated that question when i said \[r=\sqrt{x^2+y^2}\] so if, for example the point was \((2,5)\) you would have \[r=\sqrt{2^2+5^2}\]
anonymous
  • anonymous
i notice on this page at least, the \(\theta\)'s have made an appearance !
anonymous
  • anonymous
Lol. Would i leave it |dw:1440642870108:dw|
anonymous
  • anonymous
keep the radical
anonymous
  • anonymous
ok
anonymous
  • anonymous
notice it says "no decimal"
anonymous
  • anonymous
yeah. i just seen that
anonymous
  • anonymous
So #7 would be |dw:1440643014145:dw|
anonymous
  • anonymous
nvm
anonymous
  • anonymous
lol
anonymous
  • anonymous
|dw:1440643104712:dw|
anonymous
  • anonymous
it is really \[\sin(\theta)=\frac{5}{\sqrt{29}}\] but that is not what they want as a "final answer"q
anonymous
  • anonymous
yeah cause i cant leave the 29 like that.
anonymous
  • anonymous
I would have to multiply ???
anonymous
  • anonymous
aka "rationalize the denominator "
anonymous
  • anonymous
\[\frac{5}{\sqrt{29}}\times \frac{\sqrt{29}}{\sqrt{29}}=\frac{5\sqrt{29}}{29}\] with a tiny bit of practice you do this in your head
anonymous
  • anonymous
|dw:1440643207188:dw|
anonymous
  • anonymous
there should be a 29 in the denominator
anonymous
  • anonymous
So thats r??
anonymous
  • anonymous
no \(r=\sqrt{29}\)
anonymous
  • anonymous
\[\sin(\theta)=\frac{5\sqrt{29}}{29}\]
anonymous
  • anonymous
hmmm so cos (theta)= 2/29?
anonymous
  • anonymous
no \[\cos(\theta)=\frac{2}{\sqrt{29}}=\frac{2\sqrt{29}}{29}\]
anonymous
  • anonymous
Tan = 5/2??
anonymous
  • anonymous
yes
anonymous
  • anonymous
nice. im doing #8 now
anonymous
  • anonymous
ok find r as before, then use the same ratios as before
anonymous
  • anonymous
ok
anonymous
  • anonymous
the negatives are messing me up when im using the ratios
anonymous
  • anonymous
Am i suppose to us positives??
anonymous
  • anonymous
no
triciaal
  • triciaal
plot the points and pay attention to the signs
anonymous
  • anonymous
leave them as negative
anonymous
  • anonymous
the tangent will be positive since you have two negatives, but the sine and cosine are negative for sure
triciaal
  • triciaal
|dw:1440644210522:dw|
anonymous
  • anonymous
So is this correct????
triciaal
  • triciaal
|dw:1440644481593:dw|
triciaal
  • triciaal
correct
anonymous
  • anonymous
I got it correct??
triciaal
  • triciaal
now you have the CAST rule
triciaal
  • triciaal
|dw:1440644731475:dw|
triciaal
  • triciaal
another mnemonic to remember what ratios are positive in which quadrant
anonymous
  • anonymous
ok thanks
triciaal
  • triciaal
A for all positive in quadrant 1
triciaal
  • triciaal
you are welcome
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
@triciaal What about the S T C
anonymous
  • anonymous
anonymous
  • anonymous
@satellite73

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