Question

Does anyone know how I can find the quadrant of A-B?

Answer 1

No, sorry I don't

Answer 2

quadrant of A-B?

Answer 3

I was told that given, Tan(A) = 5 in Q3 and Sin(B) = 2/3 in Q2, to find Sin(A-B). I got (5√(130) + 2√26)/78.
Then I had to find Cos(A-B), and I got (-1/26)(-√5/3) + (-5/√26)(2/3).
Now I'm supposed to find what quadrant A-B is in.

Answer 4

oh

Answer 5

@freckles yep. It says "what is the Quadrant of A-B?"

Answer 6

(x,y)=(cos(A-B),sin(A-B))

Answer 7

if (x,y)=(+,+) then Q1
if (x,y)=(-,+) then Q2
if (x,y)=(-,-) then Q3
if (x,y)=(+,-) then Q4

Answer 8

@freckles so it would be Q1 then?

Answer 9

if sin(A-B) and cos(A-B) are both positive outputs yes

Answer 10

do you want me to also check your work for sin(A-B) and cos(A-B)?

Answer 11

@freckles yes, I would appreciate it!

Answer 12

i got a different sign on one of the terms in your numerator for the first one

Answer 13

@freckles what did you get?

Answer 14

actually nevermind i think what you have is right
and that one expression I think you just left off the sqrt on the 26

Answer 15

are you sure both sin(A-B) and cos(A-B) are positive

you might want to approximate them to see

Answer 16

@freckles oh I see. My mistake there. Thank you for helping me on this, and for checking my answers :)

Answer 17

@freckles ok, how should I go about approximating them?

Answer 18

calculator :p

Answer 19

punch numbers in

Answer 20

@freckles looks like Sin(a-b) is positive and Cos(a-b) is negative, so it would be in the 2nd Q, yes?

Answer 21

yes

Answer 22

@freckles got it! thanks again!

Answer 23

np

Answer 24

if you combine terms, you can get to
\[ \sin A-B = \frac{2+ 5\sqrt{5}}{3 \sqrt{26}} \]which is positive
and
\[ \cos A-B= \frac{\sqrt{5}-10} {3 \sqrt{26}} \]
the sqr(5) is a bit more than 2 (definitely not 3 because 3*3=9(
so the top is negative , and the bottom is positive.

so you have sin + and cos -


I always translate sin to "y" and cos to "x"
so -x, + y is the 2nd quadrant