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IF A+B,A ARE ACUTE ANGLES SUCH THAT SINA+B =24/25 AND TAN A =3/4FIND COSB

IF A+B,A ARE ACUTE ANGLES SUCH THAT 
SINA+B =24/25 AND TAN A =3/4FIND COSB
 

Grade:11

1 Answers

Vikas TU
14149 Points
6 years ago
We know the formulae as:
sin(A + B) = sin(A)cos(B) + sin(B)cos(A) 

If tan A = 3/4, then the opposite side is 3, and the adjacent side is 4, so the hypotenuse is 5. 

Therefore sin A = 3/5 and cos A = 4/5 
sin(B) = √1 - cos²(B) 
sin(A + B) = sin(A)cos(B) + sin(B)cos(A) 
24/25 = 3/5cos(B) + 4/5sin(B) 
24/25 = 3/5cos(B) + 4/5√1 - cos²(B)
24/5 = 3cos(B) + 4√1 - cos²(B) 
24/5 - 3cos(B) = 4√1 - cos²(B) 
24 - 15cos(B) = 20√1 - cos²(B) 

Square both sides. 

576 - 720cos(B) + 225cos²(B) = 400 - 400cos²(B) 

625cos²(B) - 720cos(B) + 176= 0  
-(-720) ±√(-720)² - 4(625)(176) 
--------------------------------------... = cos(B) 
2(625) 

720 ±√78400 /1250
= cos(B)

cos(B) = .352 or cos(B) = .8

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