Explain SOH-CAH-TOA
sin=______
cos=______
tan=_______

- anonymous

Explain SOH-CAH-TOA
sin=______
cos=______
tan=_______

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- anonymous

@peachpi

- triciaal

|dw:1440640439967:dw|

- triciaal

|dw:1440640548365:dw|

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## More answers

- anonymous

O thanks :)

- 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

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

@triciaal

- 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

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- anonymous

i kind of dont understand

- anonymous

@satellite73

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- 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

ok

- anonymous

i just dont know what to put in the blanks in number 5

- 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

whoever wrote this made a very serious error however

- anonymous

what was the error? It was my teacher btw lol

- 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

you can tell him/her satellite said they were abusing notation which will lead to massive confusion for you down the road

- anonymous

Oh, ill be sure to tell her. What would i label the side lying on the x axix and the vertical side??

- anonymous

x and y i guess

- anonymous

|dw:1440642321127:dw|

- anonymous

|dw:1440642400091:dw|

- anonymous

|dw:1440642460037:dw|

- anonymous

is that it?

- anonymous

yeah looks good

- 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

all right. Thanks. can you guide me with this?

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- triciaal

#6 use pythagoren theorem @satellite73 has it above

- 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

i notice on this page at least, the \(\theta\)'s have made an appearance !

- anonymous

Lol. Would i leave it |dw:1440642870108:dw|

- anonymous

keep the radical

- anonymous

ok

- anonymous

notice it says "no decimal"

- anonymous

yeah. i just seen that

- anonymous

So #7 would be |dw:1440643014145:dw|

- anonymous

nvm

- anonymous

lol

- anonymous

|dw:1440643104712:dw|

- anonymous

it is really \[\sin(\theta)=\frac{5}{\sqrt{29}}\] but that is not what they want as a "final answer"q

- anonymous

yeah cause i cant leave the 29 like that.

- anonymous

I would have to multiply ???

- anonymous

aka "rationalize the denominator "

- 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

|dw:1440643207188:dw|

- anonymous

there should be a 29 in the denominator

- anonymous

So thats r??

- anonymous

no \(r=\sqrt{29}\)

- anonymous

\[\sin(\theta)=\frac{5\sqrt{29}}{29}\]

- anonymous

hmmm so cos (theta)= 2/29?

- anonymous

no \[\cos(\theta)=\frac{2}{\sqrt{29}}=\frac{2\sqrt{29}}{29}\]

- anonymous

Tan = 5/2??

- anonymous

yes

- anonymous

nice. im doing #8 now

- anonymous

ok find r as before, then use the same ratios as before

- anonymous

ok

- anonymous

the negatives are messing me up when im using the ratios

- anonymous

Am i suppose to us positives??

- anonymous

no

- triciaal

plot the points and pay attention to the signs

- anonymous

leave them as negative

- anonymous

the tangent will be positive since you have two negatives, but the sine and cosine are negative for sure

- triciaal

|dw:1440644210522:dw|

- anonymous

So is this correct????

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- triciaal

|dw:1440644481593:dw|

- triciaal

correct

- anonymous

I got it correct??

- triciaal

now you have the CAST rule

- triciaal

|dw:1440644731475:dw|

- triciaal

another mnemonic to remember what ratios are positive in which quadrant

- anonymous

ok thanks

- triciaal

A for all positive in quadrant 1

- triciaal

you are welcome

- anonymous

@satellite73

- anonymous

@triciaal What about the S T C

- anonymous

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- anonymous

@satellite73

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