If a triangle ABC is right-angled at A, then what is the value of $sin\frac{B+C}{2}cos\frac{B+C}{2}$ ?
Answer & explanation
Correct answer: option 4
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 \(\frac{B+C}{2}\)
= \(\frac{1 }{√2}\). \(\frac{1 }{√2}\)
= \(\frac{1 }{2}\)