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

Nikhil Rastogi
25 Points
3 years ago
$Sin^2x = 1- cos^ 2x$
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
20 Points
2 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
13 Points
2 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
16 Points
2 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