Question

in a triangle A,B,C if sin2A+2B=sin2C. prove that A=90 orB=90

Answer 1

Solution

We have:sin2A=sin2B+sin2C

⇒(sin2B+sin2C)−sin2A=0

⇒2sin(2B+2C2)cos(2B−2C2)−2sinA.cosA=0

⇒2sin(B+C)/2 cos(B−C)−2sinA

.cosA=0Dividing both sides by 2,

 we get,⇒sin(B+C)cos(B−C)/2

−2sinA.cos =0Now

 A,B,C are angles of the triangle

, henceA+B+C=180°

⇒sin(B+C)cos(B−C)−sin{180°−(B+C)}
.cos{180°−(B+C)}=0

⇒sin(B+C)cos(B−C)−sin(B+C).{−cos(B+C)}=0

⇒sin(B+C)cos(B−C)+sin(B+C).cos(B+C)=0

Dividing both sides by sin(B+C), 

.we get

,cos(B−C)+cos(B+C)=0

⇒cos(B+C)+cos(B−C)=0

⇒2cos(B+C+B−C2)cos(B+C−B+C2)=0

⇒cos(B).cos(C)=0Hence

.Either cos(B)=0 or cos(C)=0As cos(90°)=0

⇒B=90° or C=90°  [Hence proved]