A and B are positive acute angles. If sin A=(3/5) and cosB= (12/13) find the value of sin (A+B)
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first you have to find the angles cosA and
SinB using pythogoras theorem. Then using the property expand Sin(A+B)and then substitute the given values. similarly do the same for cos(A+B) and find the quadrant in which A+B lies. That's it!!!
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obviously its math so can you please just do the problem
To find the value of sin(A + B), we can use the trigonometric identity:
sin(A + B) = sin A * cos B + cos A * sin B
Given that sin A = 3/5 and cos B = 12/13, we first need to find cos A and sin B.
Using the Pythagorean identity:
sin^2 A + cos^2 A = 1
We can substitute sin A = 3/5, and solve for cos A:
(3/5)^2 + cos^2 A = 1
9/25 + cos^2 A = 1
cos^2 A = 1 - 9/25
cos^2 A = 25/25 - 9/25
cos^2 A = 16/25
Taking the square root of both sides:
cos A = ±4/5
Since angle A is positive and acute, we take the positive value:
cos A = 4/5
Now, let's find sin B. Since cos B = 12/13, we can use the Pythagorean identity:
sin^2 B = 1 - cos^2 B
sin^2 B = 1 - (12/13)^2
sin^2 B = 1 - 144/169
sin^2 B = 169/169 - 144/169
sin^2 B = 25/169
Taking the square root of both sides:
sin B = ±5/13
Since angle B is positive and acute, we take the positive value:
sin B = 5/13
Now we can substitute the values of sin A, sin B, cos A, and cos B into the formula for sin(A + B):
sin(A + B) = sin A * cos B + cos A * sin B
sin(A + B) = (3/5) * (12/13) + (4/5) * (5/13)
sin(A + B) = 36/65 + 20/65
sin(A + B) = 56/65
Therefore, the value of sin(A + B) is 56/65.