Q. 7

# In a right-angled triangle ABC, write the value of sin^{2} A + sin^{2} B + sin^{2} C.

Answer :

Given, triangle ABC is right angle.

So, let ∠ B = 90°

Then as per the property of angles in a triangle

∠ A + ∠ B + ∠ C = 180°

As ∠ B = 90°

∠ A + 90° + ∠ C = 180°

Then ∠ A + ∠ C = 180° - 90° = 90°

Now, consider sin ^{2}A + sin ^{2}B + sin ^{2}C

As ∠ B = 90°

sin^{2}A + sin^{2}B + sin^{2}C = sin^{2}A + sin^{2}(90°) + sin^{2}C

= sin^{2}A + 1 + sin^{2}C

From before, we know that ∠ A + ∠ C = 90° ; ∠ C = 90° - ∠ A

sin^{2}A + sin^{2}B + sin^{2}C = sin^{2}A + 1 + sin^{2}( 90° - A)

= sin^{2}A + cos^{2}(A) + 1

[by using the identity cos x = sin ( 90° - x)]

sin^{2}A + sin^{2}B + sin^{2}C = (sin^{2}A + cos^{2}A) + 1

= 1 + 1

= 2

[by using the identity sin^{2}θ + cos^{2}θ = 1]

Therefore, sin^{2}A + sin^{2}B + sin^{2}C = 2.

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