if in an acute angled triangle abc if sin A+B-C = 1/2 and cos B+C- A = 1/root2 find angles a b c
Answers
Angles of ‘A’, ‘B’ and ‘C’ are
GIVEN:
A+B+C = 1/2
To find:
Angles of A,B nd C
Solution:
Let us take the triangle ABC, now as the question says
Therefore, the values can be written, as
and for the cosine the value is
So, as we know that the sum of all ‘sides of a triangle’ is 180,
Hence,
With all the equation, we can find the values of A, B and C. Equating all equation:
Let us take
Substituting value of A+B=C+30 in A+B+C=180, we get
Now with we can find the value of A and B, putting the value of C in A+B=C+30 we get
And putting the value of we get B+75-A=45; B-A=-30 or A-B=30
Equate A-B=30 & A+B=105 we get the value of A as
Therefore, value of B is
Hence, the angles of ‘A’, ‘B’ and ‘C’ are
A = 67.5°
B = 37.5°
C = 75°
Given:
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°