Question

Can someone please check my work? (see below)

Answer 1

not done...

Answer 2

\[36^2+24^2=1872\]
hypotenuse = \[\sqrt{1872}\]

Answer 3

still not done

Answer 4

SOHCAHTOA !

Answer 5

\[\sin(A)=\frac{24}{\sqrt{1872}};\cos(A)=\frac{36}{\sqrt{1872}};\tan(A)=\frac{24}{36}\]
\[\sin(A)=\frac{24\sqrt{1872}}{1872}; \cos(A)=\frac{36\sqrt{1872}}{1872}; \tan(A)=\frac{2}{3}\]

Answer 6

still not done...

Answer 7

AB = 43.267

Answer 8

\[\sin(A)=\frac{24\sqrt{144(13)}}{144(13)};\sin(A)=\frac{36\sqrt{144(13)}}{144(13)};\]
\[\sin(A)=\frac{24(12)\sqrt{13}}{144(13)};\cos(A)=\frac{36(12)\sqrt{13}}{144(13)}\]

Answer 9

still not done

Answer 10

\[\sin(A) = \frac{288\sqrt{13}}{144(13)};\;cos(A) = \frac{432\sqrt{13}}{144(13)}\]
\[\sin(A)=\frac{2\sqrt{13}}{13};\cos(A)=\frac{3\sqrt{13}}{13};\]

Answer 11

^^Sorry, up there typo. It should read cos(A) = 36sqrt{144(13)}...etc

Answer 12

Well....does this look right?

Answer 13

@zepdrix

Answer 14

YES

Answer 15

tan A = 2/3
sinA / cosA = 2 sqrt13/13 * 13/ 3sqrt13 = 2/3

Answer 16

So...^ that indicates I'm correct, right?

Answer 17

Gurrrrl :)
You should probably simplify your hypotenuse BEFORE you use it.
Otherwise you gotta do all of this extra simplification every time.\[\Large\rm \sqrt{1872}=12\sqrt{13}\]It's just a lot easier to carry this number around, isn't it? :d

Answer 18

OMG. How stupid of me...

Answer 19

How did I not think of that?

Answer 20

I don't see any mistakes yet..
just looks like you gave yourself some extra work though hehe

Answer 21

the algebra is correct nontheless

Answer 22

lol thx you two. :)

Answer 23

- well i cant see anything wrong...