Question

In a triangle ABC if sin2A=sin2B+sin2C then find measures of angles A ,B and C

Answer 1

Answer:

A=0° or B =90° or C=90°

Step-by-step explanation:

Given ---> In a ΔABC if

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Sin2A =Sin2B +Sin2C

To find ---> Measure of angles A,B and C

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Solution--->

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Sin2A= Sin2B + Sin 2C

=> Sin2B + Sin2C = Sin2A

we know that

Sinx + Siny =2Sin (x+y/2) Cos(x-y/2)

Applying it we get

=>2Sin(2B+2C/2) Cos(2B-2C/2)- Sin2A=0

=>2Sin(B+C) Cos(B-C) - Sin2A =0

A + B + C= π

=> B + C =π -A

=> 2 Sin (π-A) Cos(B-C) - Sin2A=0

=> 2SinA Cos(B-C) - 2SinA CosA=0

=> 2SinA { Cos(B-C) - CosA }=0

=> 2SinA [Cos(B-C) - Cos{π-(B+C)}] =0

=>2 SinA {Cos(B-C) + Cos (B+C)} =0

We have a formula

Cosx+ Cosy=2Cos(x+y/2) Cos(x-y/2)

Here x =B+C and y =B-C

x+y = B+C +B-C =2B

x-y=B+C-B+C=2C

So applying this formula

=>2 SinA 2 Cos(2B/2) Cos(2C/2) =0

=>4 SinA CosB CosC = 0

If SinA = 0

SinA =Sin0°

A = 0°

If CosB=0

CosB=Cos90°

B= 90°

If CosC=0

CosC=Cos90°

C=90°

Answer 2

Answer:

Step-by-step explanation:

SOLUTION...