An exterior angle at the base of an isosceles triangle is always Obtuse.
Let \(\Delta \rm ABC\) is an isosceles triangle having \(\angle \rm B\)\(=\angle\rm C\).
We know that, the sum of all the interior angle of triangle is \(180^\circ\).
So, \(\angle\rm A+\angle \rm B+\angle\rm C=180^\circ\)
\(\angle \rm A+\angle B+\angle B=180^\circ\) ( Since, \(\angle \rm B=\angle\rm C\) ).
\(\angle \rm A+2\angle B=180^\circ\)
\(\angle \rm A=180^\circ-2\angle \rm B\).......(1).
Since, An exterior angle of a triangle is equal to the sum of its two opposite non-adjacent interior angles.
So, exterior angle C,
\(\rm Exterior\angle \rm C=\angle A+\angle B\)
Putting the value of \(\angle\rm A\) from equation (1), we get
\(\rm Exterior\angle \rm C=180^\circ-2\angle \rm B+\angle B\)
\(\rm Exterior\angle \rm C=180^\circ-\angle B\)......(2)
Now from equation (2), we make a conclusion about the type of \(\rm Exterior\angle \rm C\).
If the value of \(\angle \rm B\) exceeds 90 degrees than only the \(\rm Exterior\angle \rm C\) value will be less than 90 degrees,
but in \(\Delta \rm ABC\) , \(\angle \rm B=\angle\rm C\) so it is not possible that the value of angle B will be greater than 90 degrees.
Hence we can say that in every case the value of \(\rm Exterior\angle \rm C\) will be always greater than 90 degrees, that is why we say that the \(\rm Exterior\angle \rm C\) is always obtuse.
Thus, the correct option is 1. Obtuse.
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