In a triangle ABC if sin2A=sin2B+sin2C then find measures of angles A ,B and C
Answers
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:
Step-by-step explanation:
SOLUTION...
