Question

In a ∆ABC right angled at B, ∠A = ∠C. Find the values of
(i) sin A cos C + cos A sin C
(ii) sin A sin B + cos A cos B

Answer 1

SOLUTION :  

(i) Given : In a Δ ABC right angled ∆ at B and ∠A = ∠C.

The value of ∠B is 90° (Given)

In ∆ ABC , ∠B = 90°  

∠A +  ∠B + ∠C = 180°

[sum of angles of triangle is 180∘]

∠A +  90° + ∠A = 180°

[∠A = ∠C]

2∠A  + 90° = 180°

2∠A  = 180° - 90°

2∠A = 90°

∠A = 90° /2

∠A = 45°

Hence, the value of ∠A = ∠C is 45°.

sin A cos C + cos A sin C

sin 45° cos 45° + cos 45° sin 45°

[On putting the value of ∠A = ∠C = 45°]

= (1/√2 ×1/√2) + (1/√2 × 1/√2)

[sin 45° = 1/√2 , cos 45° = 1/√2]

= 1/2 + 1/2

= (1 + 1)/2

= 2/2 = 1

sin A cos C + cos A sin C = 1

Hence,the value of  sin A cos C + cos A sin C = 1

(ii)  sin A sin B + cos A cos B

sin 45°  sin 90° + cos 45° cos 90°

= 1/√2 × 1 + 1/√2 × 0

[sin 45° = 1/√2 , cos 45° = 1/√2,sin 90° = 1, cos 90°  =0]

= 1/√2 + 0

=  1/√2

sin 45°  sin 90° + cos 45° sin 90° = 1/√2

Hence, the value of sin A sin B + cos A cos B = 1/√2.

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

hello....


 ∠A + ∠B + ∠C = 180°

⇒ ∠A + 90° + ∠C = 180°

⇒ 2∠A  = 90° 

⇒ ∠A = ∠C = 45°

Cos A = Cos 45° =  1/√2


 Cos C = Sin 45° =    1/√2

Now, Sin B = Sin 90° = 1

Cos B = Cos 90° = 0


i) Sin A . Cos C  + Cos A . Sin C

=   1/√2 ×   1/√2 +   1/√2 ×   1/√2 =    1/√2 +  1/√2= 1



ii) Sin A . SinB + Cos A . Cos B 

=   1/√2  × 1 +   1/√2 × 0 =   1/√2


thank you..