Math, asked by devasahu2669, 1 year ago

In an acute angled triangle ABC, if sin(A+B-C) = 1/2 and cos(B+C-A) = 1/root2 find angleA, angleB, angleC

Answers

Answered by Zico26
1
sin(A+B-C)=sin30
or,a+b-c=30
cos(b+c-a)=0
or,b+c-a=0
now solve it
Hope it helps u
Thank u

jyoti174: its in wrong method
Answered by adarshnayak130
1

step-by-step explanation:

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°

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°

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