Sin (a+b-c)=1/2 and cos ( b+c-a)= 1/sqrt 2 find a , b and c
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In an acute angled triangle ABC,if ...
Homework Help > CBSE > Class 10 > Mathematics
In an acute angled triangle ABC,if sin (A+B-C)= 1/2 and cos (B+C-A)= 1/root 2
Posted by Sweety Agarwal 1 year, 1 month ago
CBSE > Class 10 > Mathematics
1 answers
Manish Gandhi1 year, 1 month ago
Sin (A+B-C) = 1/2 => Sin (A+B-C) = Sin 300
Cos (B+C-A) = 1/√2 => Cos (B+C-A) = Cos 450
=> A + B - C = 300 ........ (Eq. 1)
AND, B + C - A = 450 ........ (Eq. 2)
ALSO, A + B + C = 1800 ........ (Eq. 3)
Adding Eq. 1 & 2, we get: 2 B = 750 => B = 37.50
Subtracting eq. 2 from eq. 3, we get: 2 A = 1350 => A = 67.50
Using value of A & B, in Eq. 3, we get C = 750
In an acute angled triangle ABC,if ...
Homework Help > CBSE > Class 10 > Mathematics
In an acute angled triangle ABC,if sin (A+B-C)= 1/2 and cos (B+C-A)= 1/root 2
Posted by Sweety Agarwal 1 year, 1 month ago
CBSE > Class 10 > Mathematics
1 answers
Manish Gandhi1 year, 1 month ago
Sin (A+B-C) = 1/2 => Sin (A+B-C) = Sin 300
Cos (B+C-A) = 1/√2 => Cos (B+C-A) = Cos 450
=> A + B - C = 300 ........ (Eq. 1)
AND, B + C - A = 450 ........ (Eq. 2)
ALSO, A + B + C = 1800 ........ (Eq. 3)
Adding Eq. 1 & 2, we get: 2 B = 750 => B = 37.50
Subtracting eq. 2 from eq. 3, we get: 2 A = 1350 => A = 67.50
Using value of A & B, in Eq. 3, we get C = 750
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