Question

ABC is a right angle triangle then find the value of sinA+sinB+sinC​

Answer 1

Question:-

ABC is a right angle triangle then find the value of sinA+sinB+sinC

${\red{\underline{\underline{\bold{Answer:-}}}}}$

Since ∠ A = ∠ C

It must be an Isosceles Triangle.

So, ∠ A + ∠ B + ∠ C = 180°

= 2 ∠ A + 90° = 180°

= 2 ∠ A = 180° - 90° = 90°

= ∠ A = 90° ÷ 2 = 45 °

=> ∠ A  = ∠ C = 45°

So Sin A = Sin 45 = 1 / √2

Sin B = Sin 90 = 1

So Sin A × Sin B = 1 / √2 × 1

= 1 /√2

Cos A = Cos 45 = 1 / √2

Cos B = Cos 90 = 0

So Cos A × Cos B = 1 / √2 × 0

= 0

So Sin A × Sin B + CosA × Cos B = 1 / √2 + 0

= 1 / √2

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