Question

HELP ME PLEASE I NEED THIS TEST DONE ASAP!!!

Answer 1

Ok so triangle A. Ur given that 2/3 is cos so that means that u have ur adjacent and hypotenuse 2 being the adj and 3 the hypotenuse. So u use the Pythagorean theorem formula to find ur opposite 2^2 +b^2= 3^2
4+b^2=9 move the 4 around
b^2= 5 take the square root and it’s just radical 5
So sine in this case for triangle A is radical 5 over 3

Now ur triangle B ur given cos again so -3 is ur adjacent and 4 the hypotenuse. So u use the same formula so
(-3)^2+b^2=4^2
9+b^2=16 u move the 9
b^2=7 take the square root so it’s just radical 7
Meaning that ur sine is for triangle B is radical 7 over 4

Now for sin(a-b)
U take the sine of triangle A and B so
Radical 5 over 3 - radical 7 over 4
U must have a common denominator so multiply radical 5 over 3 by 4 top and bottom and radical 7 over 4 by 3 top and bottom.
4radical 5/12 - 3radical7/12
Simplified is 0.08
So Sin(0.08)=0.001

And for cos u already know the cos of both triangles so u just find the common denominator
2/3 - -3/4
4/12 - -9/12
1.08
Cos(1.08)
0.9999

I hope I’m correct if not I’m sorry I tried :(