If a triangle ABC is right-angled at A, then what is the value of $sin\frac{B+C}{2}cos\frac{B+C}{2}$ ?

$\frac{\sqrt{3}}{4}$ $\sqrt{2}$ $\frac{1}{4}$ $\frac{1}{2}$

$\frac{1}{2}$

Triangle  is right angled at A . So, sum of other 2 angles are 90º. So, \(\frac{B+C}{2}\) = \(\frac{90º}{2}\) = 45º Now, sin \(\frac{B+C}{2}\) . cos \(\frac{B+C}{2}\) = \(\frac{1 }{√2}\). \(\frac{1 }{√2}\) = \(\frac{1 }{2}\)