Question

If triangle ABC is a right angled at C prove that : sin B + cos B > 1

Answer 1

By sine rule,
a/sinA=b/sinB=c/sinC=2R

Now we know that in triangle, (a+b>c)-----(1)
a can be written as 2RsinA and b can be as 2RsinB and c as 2RsinC
Putting values in equation 1,
2R(sinA+sinB)>2RsinC
Now A can be written as (180-(B+C))and we know that sin(180-#)=sin# so sinA=sin(B+C) and angle C=90which makes it sin(90+B) i.e.cosB
So we finally get that,,,,,,,, (cosB+SinB) >1
We have put the value of C in sinC