Question

In an acute angle triangle ABC, if cos(A+B-C)=1/root2 and sin(B+C-A)=root3/2, then find the value of A, B, C. Please help its urgent, if correct answer will mark brainlist and 1000 thanks

Answer 1

Answer:

Triangle is acute angle angled triangle

. Also given that, cos(a+b-c) = 1 √ 2 12 = cos 45° And, sin(b+c-a) = √ 3 2 32 = sin 60° ∴ a+b-c = 45° ....(1) And, b+c-a = 60° ....(2)

By adding (1) & (2) we get, 2b = 105° ⇒ b = 105 ° 2 105°2= 52.5°

Also, a+b+c = 180° ...(3) (∵ sums of angles is 180°) By (3) - (1) we get, 2c = 135° ⇒ c = 135 ° 2 135°2 = 67.5° By (3) - (2) we get, 2a = 120° ⇒ a = 60°