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If A, B and C are interior angles of a triangle ABC, then show that sin (B+C/2) = cos A/2

Harshit Singh , 3 Years ago
Grade 12th pass
anser 1 Answers
Pawan Prajapati

Last Activity: 3 Years ago

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180° A + B + C = 180° ….(1) To find the value of (B+ C)/2, simplify the equation (1) ⇒ B + C = 180° – A ⇒ (B+C)/2 = (180°-A)/2 ⇒ (B+C)/2 = (90°-A/2) Now, multiply both sides by sin functions, we get ⇒ sin (B+C)/2 = sin (90°-A/2) Since sin (90°-A/2) = cos A/2, the above equation is equal to sin (B+C)/2 = cos A/2 Hence proved.

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