 Hi and welcome to the session. I am Shrashi and I am going to help you with the following question. If A, B and C are interior angles of a triangle ABC then show that sin B plus C upon 2 is equal to cos A upon 2. First of all let us understand that if theta is any acute angle in any right triangle then sin 90 minus theta is equal to cos theta. This is the key idea to solve the given question. Now let us start the solution. We are given that A, B and C are interior angles of a triangle ABC. So we can write in triangle ABC A plus B plus C is equal to 180 degrees by angle some property of a triangle subtracting A from both the sides we get B plus C is equal to 180 degrees minus A. Now dividing both the sides by 2 we get E plus C upon 2 is equal to 90 degrees minus A upon 2. Now clearly we can see these two angles are equal so their sin ratio will also be equal to each other. So we can write sin B plus C upon 2 is equal to sin 90 degrees minus A upon 2. We know sin 90 minus theta is equal to cos theta. Here value of theta is A upon 2. So sin 90 minus A upon 2 is equal to cos A upon 2. So we get sin B plus C upon 2 is equal to cos A upon 2. So our required answer is sin B plus C upon 2 is equal to cos A upon 2. Hence proved. This completes the session. Hope you understood the solution. Take care and keep smiling.