Saved Bookmarks
| 1. |
If A, B and C are interior angles of a triangle ABC, then show that sin [ B + C / 2 ] = cos A/2. |
|
Answer» Then proved that sin[B+C/2]=cosA/2 And sin(90°-A/2)= CosA/2 Dividing both side by 2 ||. B+C/2=90°-A/2 We know that,A+B+C=180° ||. B+C=180°-A ??? |
|