Question

pls solve this question​

Answer 1

Given :

A = 67.5°

B = 37.5°

C = 75°

To find :

$\pink\implies \sf\sin \: (A+B-C)=\frac{1}{2}$

$\pink\implies \sf\cos \: (B+C-A)=\frac{1}{\sqrt{2}}$

Solution :

We know that in a triangle, sum of the angles = 180°

A + B + C = 180 → (1)

We know that,

So,

sin (A + B - C) = sin 30

A + B - C = 30 → (2)

And

cos (B + C - A) = cos 45

B + C - A = 45 → (3)

On solving equation (1) and (2), we get,

A + B + C - A - B + C = 180 - 30 = 150

2C = 150

C = - 75 = 30

A + B = 105 → (4)

Also, substituting in equation (3), we get,

B + 75 - A =45

A - B = 30 → (5)

Adding equations (4) and (5), we get,

2A = 135 → A = 67.5°

B = A - 30 = 67.5 - 30 = 37.5°

Therefore, A = 67.5°; B = 37.5°; and C = 75

【 ✌YellowVine✌ 】

Answer 2

Step-by-step explanation:

Given :

A = 67.5°

B = 37.5°

C = 75°

To find :

\pink\implies \sf\sin \: (A+B-C)=\frac{1}{2}⟹sin(A+B−C)=

2

1

\pink\implies \sf\cos \: (B+C-A)=\frac{1}{\sqrt{2}}⟹cos(B+C−A)=

2

1

Solution :

We know that in a triangle, sum of the angles = 180°

A + B + C = 180 → (1)

We know that,

So,

sin (A + B - C) = sin 30

A + B - C = 30 → (2)

And

cos (B + C - A) = cos 45

B + C - A = 45 → (3)

On solving equation (1) and (2), we get,

A + B + C - A - B + C = 180 - 30 = 150

2C = 150

C = - 75 = 30

A + B = 105 → (4)

Also, substituting in equation (3), we get,

B + 75 - A =45

A - B = 30 → (5)

Adding equations (4) and (5), we get,

2A = 135 → A = 67.5°

B = A - 30 = 67.5 - 30 = 37.5°

Therefore, A = 67.5°; B = 37.5°; and C = 75

Hope it's helpful to you