**(i)** As we have #a^2+b^2=c^2#, which means that sum of the squares of the two sides #a# and #b# is equal to square on the third side #c#. Hence, #/_C# opposite side #c# will be right angle.

Assume, it is not so, then draw a perpendicular from #A# to #BC#, let it be at #C'#. Now according to Pythagoras theorem, #a^2+b^2=(AC')^2#. Hence, #AC'=c=AC#. But this is not possible. Hence, #/_ACB# is a right angle and #Delta ABC# is a right angled triangle.

Let us recall the cosine formula for triangles, which states that #c^2=a^2+b^2-2abcosC#.

**(ii)** As range of #/_C# is #0^@< C< 180^@#, if #/_C# is obtuse #cosC# is negative and hence #c^2=a^2+b^2+2ab|cosC|#. Hence, #a^2+b^2< c^2# means #/_C# is obtuse.

Let us use Pythagoras theorem to check it and draw #DeltaABC# with #/_C>90^@# and draw #AO# perpendicular on extended #BC# as shown. Now according to Pythagoras theorem

#a^2+b^2=BC^2+AC^2#

= #(BO-OC)^2+AC^2#

= #BO^2+OC^2-2BOxxCO+AO^2+OC^2#

= #BO^2+AO^2-2OC(BO-OC)#

= #AB^2-2OCxxBC=c^2-OCxxBC#

Hence #a^2+b^2< c^2#

**(iii)** and if #/_C# is acute #cosC# is positive and hence #c^2=a^2+b^2-2ab|cosC|#. Hence, #a^2+b^2> c^2# means #/_C# is acute.

Again using Pythagoras theorem to check this, draw #DeltaABC# with #/_C<90^@# and draw #AO# perpendicular on #BC# as shown. Now according to Pythagoras theorem

#a^2+b^2=BC^2+AC^2#

= #(BO+OC)^2+AO^2+OC^2#

= #BO^2+OC^2+2BOxxCO+AO^2+OC^2#

= #AB^2+2OC(CO+OB)#

= #c^2+2axxOC#

Hence #a^2+b^2> c^2#