The sides of a triangle are \[\sin \alpha ,\,\cos \alpha \] and \[\sqrt {1 + \sin \alpha \cos \alpha } \] for some \[0 < \alpha < {\pi \over 2}\]. Then the greatest angle of the triangle is :
Solutions · 1
Official solution
Let \[a = \sin \alpha ,b = \cos \alpha \]
and \[c = \sqrt {1 + \sin \alpha \cos \alpha } \]
Clearly a and b<1 but c>1
as \[\,\,\,\sin \alpha > 0\] and \[\cos \alpha > 0\]
\[\therefore\] c is the greatest side and greatest angle is C
\[\therefore\] \[\cos C = {{{a^2} + {b^2} - {c^2}} \over {2ab}}\]
\[ = {{{{\sin }^2}\alpha + {{\cos }^2}\alpha - 1 - \sin \alpha \cos \alpha } \over {2\,\sin \alpha \cos \alpha }}\]
\[ = - {1 \over 2}\]
\[\therefore\] \[C = {120^ \circ }\]