Grade 12thTrigonometry

In triangle A is obtuse angle SinA=3/5,cosB=12/13 then sinC=

NV NAGARJUNA

8 Years agoGrade 12th

4 Answers

Nikhil Rastogi

8 Years ago

So using this we can finally get that sin 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

virendra mishra

8 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

V RAAGA PRANEETHA

7 Years ago

SinA=3/5 , cosB= 12/13

A is in second quadrant and B, C are in first quadrant

Therefore , cosA= – 4/5 cosB= 12/13

A+B+C = 180

A+B = 180 –C

sin ( A + B ) = sin ( 180 – C ) = sinC

sin C = sin ( A + B ) = sin A cos B + cos A sin B

Therefore , sin C = 16/65

Yash

7 Years ago

sinA = 3/5 so cosA = - 4/5 (since A obtuse); similarly, using a simple right-angled triangle if cosB = 12/13, then sinB = 5/13. C = 180 - (A + B) so sinC = sin[180 - (A + B)] = sin(A + B) (4 quadrants) so sinC = sinAcosB + cosAsinB = (3/5)*(12/13) + (-4/5)*(5/13) = 36/65 - 20/65 = 16/65