Question

In an acute angled triangle ABC, if sin(A+B-C)=$\frac{1}{2}$ and cos(B+C-A)=$\frac{1}{\sqrt{2}}$ Find $\angle{A}, \angle{B} and \angle{C}$.

Answer 1

In solution at many places angle is not symbolised, take care of this.


Solution : —



We know, sum of all angles in any triangle is 180°. So, sum of angle A , angle B and angle C is 180°


$\angle A + \angle B+ \angle C \: = 180 \degree \: \: \: \: \: \: \: \: \: \: - - - - - : (1)$



Given in question : -


sin( A + B - C ) = $\dfrac{1}{2}$



sin( A + B - C ) = sin30°


A + B - C = 30° ------ : ( 2 )




Given in question : -


cos( B + C - A ) = $\dfrac{1}{\sqrt{2}}$



cos( B + C - A ) = cos45° -------- : ( 3 )




Then,

Eq. 1 → A + B + C = 180°

Eq. 2 → A + B - C = 30°

Eq. 3 → B + C - A = 45°



$\: \: \: \: \: \: \: \underline{ Subtracting \: \textit{ ( 1 ) \: \& \: ( 2 ) , }}$


→ A + B + C - ( A + B - C ) = 180° - 30°

→ A + B + C - A - B + C = 150°

→ 2C = 150°

→ C = 75°




Now, putting the value of C in Eq( 1 ) and in Eq( 3 ) ,


B + C + A = 180° and B + C - A = 45°

B + 75° + A = 180 and B + 75 - A = 45°

B + A = 105° and A - B = 30°


adding both the formed ( above ) equations ,


B + A - B + A = 30° + 105°

2A = 135°

B = 67.5°



B + A = 105° ( from above )

67.5 + A = 105°

A = 37.5°


$\:$


kindly check the solution .

Answer 2

Step-by-step explanation:

We know that,

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°

Substituting C=75 in equation (2), we get,

A+B-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°