A+B=90 then sina sinb=
I assume you mean 90 degrees so right triangle and a^2 + b^2 = c^2
sin A = a/c
sin B = b/c
sin A sin B = a b/ c^2 = a b /(a^2+b^2)
To find the value of sina sinb when A + B = 90, we can use the fact that the sum of angles in a right triangle is 90 degrees.
Since sine is a trigonometric function that relates the ratio of the length of the opposite side to the length of the hypotenuse, we can express sina in terms of A and sinb in terms of B.
Let's assume that A is one of the acute angles in the right triangle, and B is the other acute angle. Then, we have:
A + B = 90
Now, let's take the sine of both sides of the equation:
sin(A + B) = sin(90)
Using the trigonometric identity for the sum of angles, we have:
sin(A)cos(B) + cos(A)sin(B) = 1
Since A and B are acute angles in a right triangle, we know that:
cos(A) = sin(B)
cos(B) = sin(A)
Substituting these values into the equation, we get:
sin(A)sin(B) + sin(A)sin(B) = 1
2sin(A)sin(B) = 1
Dividing both sides by 2, we obtain:
sin(A)sin(B) = 1/2
Therefore, when A + B = 90, sina sinb = 1/2.