The identity in question is
cos(a - b) = cos(a) cos(b) + sin(a) sin(b)
so that
cos(a - b) = 12/37 cos(a) + 3/5 sin(b)
Since both a and b lie in the first quadrant, both cos(a) and sin(b) will be positive. Then it follows from the Pythagorean identity,
cos²(x) + sin²(x) = 1,
that
cos(a) = √(1 - sin²(a)) = 4/5
and
sin(b) = √(1 - cos²(b)) = 35/37
So,
cos(a - b) = 12/37 • 4/5 + 3/5 • 35/37 = 153/185
 
        
             
        
        
        
Answer:
X=7
Step-by-step explanation:
So first add them all up
x-2+2x+x+3=4x+1
4x+1=29
4x=28
x=7
 
        
             
        
        
        
Answer:
D. y-2=-3(x-2)
Step-by-step explanation:
 
        
             
        
        
        
Answer:
 
 
Step-by-step explanation:
Product of b and 2 = 2b
According to the given description:

 
        
             
        
        
        
Answer:
 Q30-A, C  
Step-by-step explanation:
A) False, If the parallelogram VJHO was a rhombus, angle V wouldn't be supplementary to angle O
B) True, all opposite sides of a parallelogram parallel
C) False, the same reason as A
D) True, opposite sides of a parallelogram congruent
Q25-
angle BCA = 60 degrees
angle CAB = 27 degrees
angle BCT = 29 degrees
angle BMC = 58 degrees