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

Posted by Anushka Saha 6 years, 11 months ago

CBSE > Class 10 > Mathematics

4 answers

Neha Sirur 6 years, 11 months ago

Hope u ll get it

Yogita Ingle 6 years, 11 months ago

In a triangle, sum of all the interior angles
A + B + C = 180°
⇒ B + C = 180° - A
⇒ (B+C)/2 = (180°-A)/2
⇒ (B+C)/2 = (90°-A/2)
⇒ sin (B+C)/2 = sin (90°-A/2)
⇒ sin (B+C)/2 = cos A/2

Neha Sirur 6 years, 11 months ago

In a triangle abc we know that all three angles is equal to 180° so angle a +b+c=180. Now shift a on the other side i.e. b+c=180-a next divide both sides by 2 so it gives (b+c) /2= 90-a/2. further on put cos on both sides ---. Cos (b+c)/2= cos 90-a/2 and cos 90-a/2 = sin a/2. So substitute it and we get cos (b+c)/2 = sin a/2

Anushka ? 6 years, 11 months ago

Cos(B+C/2) CAN BE WRITTEN AS Sin(90-B+C/2)...by taking the L. C. M, we will get Sin(180-B+C/2) and we know that in a triangle 180-B+C/2 is equal to A . So, our answer will be SinA/2.

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