Question

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



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Answer 1

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

Answer 2

Answer:

B = 5/13 and cos A = 4/5

Now we know that A + B + C = 180°

So A + B = 180° - C

Sin(A+B) = Sin(180° - C)

=> SinA.CosB + SinB.CosA = Sin180°.CosC - SinC.Cos180°

=> (3/5)(12/13) + (5/13)(4/5) = (0)(cosC) - (SinC)(1)

=> 36/65 + 20/65 = -SinC

=> -56/65 = sinC

So sinC = -56/65

3 years ago

sinA=3/5,cosB=12/13then cosA=4/5,sinB=5/13

A+B+C=180

A+B=180-C

Sin(A+B)=Sin(180-C)

3/5*12/13+4/5*5/13=SinC

SinC=56/65