Question

Find the values of the sine, cosine, and tangent for angle A

Answer 1

i know how to do the soh cah toa thing i just need like moral support when it comes to finding the hypotenuse

Answer 2

@satellite73

Answer 3

@jim_thompson5910

Answer 4

we can do this

Answer 5

hahaha yay hi satelliteflightttt. im ready

Answer 6

first we need to find the hypotenuse which is
\[\sqrt{36^2+24^2}\]

Answer 7

i would use a calculator since these are rather big numbers
and the calculator i would use is this
http://www.wolframalpha.com/input/?i=sqrt%2836^2%2B24^2%29

Answer 8

\[12\sqrt{13}\] okay soooooo

Answer 9

or lets be a little smarter shall we?

Answer 10

24 and 36 have a common factor, namely 12
we can divide both by 12 and start with \(3,2\)
this will not effect the ratios in any way

Answer 11

it is easier to compute
\[\sqrt{3^2+2^2}=\sqrt{13}\]

Answer 12

\[\sin(A)=\frac{\text{opposite}}{\text{hypotenuse}}\]

Answer 13

you get
\[\sin(A)=\frac{2}{\sqrt{13}}\]

Answer 14

notice if we had used the bigger numbers we would get the same answer
\[\sin(A)=\frac{24}{12\sqrt{13}}=\frac{2}{\sqrt{13}}\]

Answer 15

and
\[\cos(A)=\frac{\text{adjacent}}{\text{hypotenuse}}\]

Answer 16

finally
\[\tan(A)=\frac{\text{opposite}}{\text{adjacent}}\] you good from there?

Answer 17

yes thats perfect and one more thing.. dont you end up with sin and cos having to times both by sqrt13/sqrt13 to get the radical out of the denominator and into the num

Answer 18

yeah if you have to "rationalize the denominator" then
\[\frac{2}{\sqrt{13}}=\frac{2\sqrt{13}}{13}\]

Answer 19

similarly \[\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}\]

Answer 20

tangent it just ordinary \(\frac{2}{3}\)

Answer 21

thats the exact answer i got :)

Answer 22

ok good
you good to go?

Answer 23

yeah i might need help on another one but ill make a new post if i do.. stick around<3

Answer 24

lol ok tag me or send me a message if you like