- #1
ritwik06
- 580
- 0
I have read the sine rule:
It states-->
sin A/a=sin B/b=sin C/c = 1/2R
where R is circumradius.
Now,
a=2Rsin A
b=2Rsin B
c=2Rsin C
For a triangle R is fixed.
In an obtuse angled triangle, the side opposite largest angle is the longest(geomtrically)
But the sine of an obtuse angle can be less than that of an acute angle. How is this possible? Is it possible that the side opposite obtuse angle isn't the largest? Or is the sin rul giving a wrong stand in this case??
It states-->
sin A/a=sin B/b=sin C/c = 1/2R
where R is circumradius.
Now,
a=2Rsin A
b=2Rsin B
c=2Rsin C
For a triangle R is fixed.
In an obtuse angled triangle, the side opposite largest angle is the longest(geomtrically)
But the sine of an obtuse angle can be less than that of an acute angle. How is this possible? Is it possible that the side opposite obtuse angle isn't the largest? Or is the sin rul giving a wrong stand in this case??