Question

If sin A=3/5 then find the value of SinACOSB+COSA Sin B, where angle c is 90.​

Answer 1

Answer:

sin(A)*cos(B)+cos(A)*sin(B)=1

Step-by-step explanation:

sin(A)=3/5=Length of opp. side to A / Length of hypotenuse

So, in that triangle:

Length of opp. side to A =  3

Length of hypotenuse    =  5

Length of the remaining side = 4 (since; 3,4&5 are pythagorean triplets)

Since: A+B+C=180degs (in a right triangle)

=>A+B = 180degs - C

=>sin(A+B) = sin(180degs-C)                #Applied sine ratio on both sides

=>sin(A)*cos(B)+cos(A)*sin(B)=sin(+C) #sin(A+B) expansion & sin is +ve in Q2

=>sin(A)*cos(B)+cos(A)*sin(B)=sin(90degs) #Given C=90degs

=>sin(A)*cos(B)+cos(A)*sin(B)= 1              #since sin(90degs)=1

Answer 2

Step-by-step explanation:

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