Question

in triangle ABC right angled at B,angle A =angleC find value of sinA cosC+cosA sinC​

Answer 1

Answer:

Here's your 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

Hope this helps!!!

Answer 2

Answer:

The value=1

Step-by-step explanation:

∠B=90°

∠A=∠C

Hyp= AC

Other sides= BC, AB

sinA= BC/AC

cosA=AB/AC

sinC= AB/AC

cosC= BC/AC

Using Pythagoras Theorem

AC²=BC²+AB²

sinA cosC+cosA sinC

=BC/AC*BC/AC +AB/AC*BC/AC

=BC²/AC² +AB²/AC²

=AB²+BC/AC²

=AC²/AC²

=1

Hope it is helpful

Please mark me in brainlist