Question

In ΔABC, ∠A is an obtuse angle. If sin A = 3/5, sin B = 5/13, find sin C.

Answer 1

Answer:

A is obtuse, B and C are acute

That is A lies in 2 nd quadrant.

B,C lies in 1 st quadrant.

A+B+C=180

A+B=180-C

sin(A+B)=sinC

sin A cosB+ cosA sin B

(3/5)(12/13)+(-4/5)(5/13)

=36/65-20/65

=16/65

Answer 2

Answer:

$\bold\red{sin C = \frac{16}{65}}$

Step-by-step explanation:

Given,

In ΔABC,

∠A is an obtuse angle

sin A = 3/5,

=> cos A = -4/5

(°.° A is obtuse, it lies in 2nd quadrant )

and,

sin B = 5/13,

=> cos B = 12/13

To find: sin C

We know that, in a triangle

∠A+∠B+∠C = 180

=> ∠A + ∠B = 180- ∠C

=> sin ( ∠A + ∠B ) = sin ( 180- ∠C )

=> sinA cosB + cosA sinB = sinC

=> sinC = (3/5)(12/13) + (-4/5)(5/13)

=> sin C = 36/65 - 20/65

=> sin C = 16/65