Question

In an acute angled triangle ABC sin 2(A+B-C) = 1 and tan(B + C -A) =√3, then find the values of A, B and C

Answer 1

Given

In an acute angled triangle ABC .

sin 2(A+B-C) = 1

tan(B + C -A) =√3

To proof

As given in the question

In an acute angled triangle ABC .

first solving the equation

sin 2(A+B-C) = 1

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

As we know

1 = sin90°

put this in the above equation

we get

$2 ( A + B - C ) = sin^{-1} (sin90^{\circ})$

2A + 2B - 2C = 90

A+B -C =45      ( first equation )

now solving the equation

we get

tan(B + C -A) =√3

$B + C -A =tan^{-1} \sqrt{3}$

$B + C -A =tan^{-1}(tan60^{\circ})$

B + C -A = 60   ( second equation )

As given  acute angled triangle ABC

thus

∠A + ∠ B +∠ C = 180°   ( Angle sum property of a triangle )

than  the third equation becomes

A + B + C = 180  ( third equation)

Now solve the equation

A+B -C =45

and  B + C -A = 60

Now  subtract  B + C -A = 60 from A+B -C =45

we get

(A+B -C) - (B + C -A) = 45-60

2A -2C = -15

Put this value in the  equation 2A + 2B - 2C = 90

-15 + 2B = 90

2B = 90 + 15

B = 52.5

now subtracted -A +B +C = 60 from A + B + C =180

A + B + C +A - B - C =180 - 60

2A = 120

A  = 60

Put the value of A , B  in the equation A + B + C =180

60 + 52.5 + C = 180

C = 180 - 112.5

C = 67.5

Thus  ΔABC is an acute angle triangle

therefore

∠A = 60°

∠B = 52.5°

∠C = 67.5°

Hence proved