Question

sinA=3/5,cosB=-12/13. sin(A+B)=?
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Answer 1

HEY THERE!

GIVEN :- sina=3/5 cosb=12/13.

TO SOLVE :- value of sin(a+b)

CONSTRUCTION :- First given is the value of sina so we will find the value of cosa

So sina = 3/5 => Perpendicular/ hypoteneus

So now using pithogorus trick Base = 4

now cosa = Base / hypoteneus

cosa= 4/5

Now given is the value of cosb so we will find sinb

So cosb=>12/13 => Base / hypoteneus

So using pithogorus trick Perpendicular = 5

now sinb => Perpendicular / hypoteneus => 5/13

SOLUTION :- As we know

sin(a+b) = sina cosb - cosa sinb

So now put the values of sina, cosa , sinb and cosb in the formula.

sin(a+b) = sina cosb - cosa sinb

=>sin(a+b) = 3/5 × 12/13 - 4/5 × 5/13

=> sin(a+b) = 36/65 - 4/13

=> sin(a+b) = (36 - 20 )/65

=> sin(a+b) = 16 / 65

HOPE IT HELPED YOU.

Answer 2

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