Question

if A,B,C are the angles of an acute angled triangle and cos (B+C-A)= 0, sin (C+B+A)= ✓3/2 find the values of A, B and C​

Answer 1

Answer:

Step-by-step explanation:

Note: Sin(C+B+A) = √3/2 is wrong. This is because Sin(C+B+A) = Sin 60° => A + B + C = 60° which is not correct. It should be Sin(C+A - B)

A + B + C = 180° ---- (1)

Given,                                                     Given,

Cos(B+C - A) = 0                                    Sin(C+A - B) = √3/2

=> Cos(B+C - A) = Cos90°                     => Sin(C+A - B) = Sin60°    

=> B+C - A = 90° ----- (2)                        => C + A - B = 60°  -------- (3)

(2) + (3)

B + C - A + C + A - B =  90 + 60

=>  2C = 150°

=> C = 75°

(1) - (2) we get

A + B + C - ( B + C - A) = 180 - 90

=> 2A = 90

=> A = 45°

Substituting values of A and C in (1),

A + B + C= 180

=> 45 + B + 75 = 180

=> B + 120 = 180

=> B = 60°

∠A = 45°, ∠B = 60°, ∠C = 75°