Question

If 0° <A and B < 90° such that cosA =5/13 and
sin B = 4/5 , then find
sin (A+B).
​

Answer 1

Answer:

sin(A+B) = 56/65

Step-by-step explanation:

formula: sin(A+B) = sinA.cosB + sinB.cosA

solution:

sinB = 4/5

cosB =

$\sqrt{{1 - \sin(a) }^{2}}$

Answer 2

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cosA = 5/13

As we Know 5, 12, 13 a Triplet

Hence,

sinA = 12/13

As √1 - cos²A = sinA

sinB = 4/5

As we Know that 3, 4, 5 is Triplet

√1 - sin²B = cosB

Hence,

cosB = 3/5

We Know,

sin(A+B) = sinAcosB + cosAsinB

= 12/13 × 3/5 + 5/13 × 4/5

= 36/65 + 20/65

= ( 36 + 20 ) / 65

= 56/65.