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In an acute angled triangle ABC,if sin (A+B-C)= 1/2 and cos (B+C-A)= 1/root 2

Posted by Priya Patel 8 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Manish Gandhi 8 years, 3 months 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

1Thank You

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