Question

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

Answer 1

Answer:

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

Answer 2

$\huge\boxed{\fcolorbox{yellow}{orange}{☞✿ANSWER:-}}$

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.