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Mathematics 19 Online
OpenStudy (anonymous):

0

OpenStudy (precal):

|dw:1407364906539:dw|

OpenStudy (australopithecus):

http://www.mathopenref.com/trigterminalside.html

OpenStudy (precal):

|dw:1407364927273:dw|

OpenStudy (precal):

what is cosine, do you recall?

OpenStudy (australopithecus):

Use SOH CAH TOA (read it like it is written) Sin(Θ) = Opposite/Hypothesis Cos(Θ) = Adjacent/Hypothesis Tan(Θ) = Opposite/Hypothesis

OpenStudy (precal):

use the Pythagorean thm to find out h

OpenStudy (precal):

then you have your answer

OpenStudy (australopithecus):

These terms can be rewritten to, Θ = arcsin(Opposite/Hypothesis) Θ = arccos(Adjacent/Hypothesis) Θ = arctan(Opposite/Hypothesis)

OpenStudy (precal):

we don't need those

OpenStudy (australopithecus):

One way to do this problem: Find the angle Θ Then just take cos(theta)

OpenStudy (solomonzelman):

5²+12²=13, but it looks like this doesn't apply here, because the point is diagonally from the triangle, I would understand if it was (-13,5)

OpenStudy (solomonzelman):

or if it was (5,-13)

OpenStudy (solomonzelman):

maybe, it IS 13, and not 12 ?

OpenStudy (solomonzelman):

-:(

OpenStudy (solomonzelman):

Well you can still solve it, but beware that 12 is a hypotenuse, not one of the legs.

OpenStudy (solomonzelman):

Or not, I am wrong

OpenStudy (solomonzelman):

(12,-5) Yes, I am wrong \(\Huge\color{blue}{ \bf | }\) \(\Huge\color{blue}{ \bf | }\) \(\Huge\color{blue}{ \bf | }\) \(\Huge\color{blue}{ \bf ^ \text{__ __ __} }\) \(\Huge\color{blue}{ \bf ✛ }\)\(\Huge\color{blue}{ \bf ^ \text{__ __ __} }\) \(\Huge\color{blue}{ \bf | }\) \(\Huge\color{blue}{ \bf | }\) \(\Huge\color{blue}{ \bf • }\)(12,-5) \(\Huge\color{blue}{ \bf | }\)

OpenStudy (solomonzelman):

GOod I am sorry, my brain wasn't working, the hyp IS 13.

OpenStudy (solomonzelman):

Follow their direction, sorry for confusing you again

OpenStudy (solomonzelman):

omg ?

OpenStudy (precal):

|dw:1407412026392:dw|

OpenStudy (precal):

|dw:1407412039376:dw|

OpenStudy (precal):

cosine (theta) = adj/hy cos(theta)=12/13

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