Question

triangle ABC, angle C = 90 degree ,then the value of sin2A+sin2B is

Answer 1

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solution:--

sin2A=(BC/AB)2

and sin2B=(AC/AB)2

So, sin2A + Sin2B

=(BC/AB)2 + (AC/AB)2

=BC2 + AC2 / AB2 (Using pythagoras theorem we get BC2 + AC2 = AB2 )

=AB2/AB2

=1

Answer 2

${ \sin }^{2} a + { \sin}^{2} (90 - a) \\ { \sin }^{2} a + { \cos}^{2} a \\ 1$
[here ABC is a triangle so A+B+C=180
A+B+90=180
A+B=90
B=90-A]
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