An exterior angle at the base of an isosceles triangle is always Obtuse.
Let [tex]\Delta \rm ABC[/tex] is an isosceles triangle having [tex]\angle \rm B[/tex][tex]=\angle\rm C[/tex].
We know that, the sum of all the interior angle of triangle is [tex]180^\circ[/tex].
So, [tex]\angle\rm A+\angle \rm B+\angle\rm C=180^\circ[/tex]
[tex]\angle \rm A+\angle B+\angle B=180^\circ[/tex] ( Since, [tex]\angle \rm B=\angle\rm C[/tex] ).
[tex]\angle \rm A+2\angle B=180^\circ[/tex]
[tex]\angle \rm A=180^\circ-2\angle \rm B[/tex].......(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,
[tex]\rm Exterior\angle \rm C=\angle A+\angle B[/tex]
Putting the value of [tex]\angle\rm A[/tex] from equation (1), we get
[tex]\rm Exterior\angle \rm C=180^\circ-2\angle \rm B+\angle B[/tex]
[tex]\rm Exterior\angle \rm C=180^\circ-\angle B[/tex]......(2)
Now from equation (2), we make a conclusion about the type of [tex]\rm Exterior\angle \rm C[/tex].
If the value of [tex]\angle \rm B[/tex] exceeds 90 degrees than only the [tex]\rm Exterior\angle \rm C[/tex] value will be less than 90 degrees,
but in [tex]\Delta \rm ABC[/tex] , [tex]\angle \rm B=\angle\rm C[/tex] 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 [tex]\rm Exterior\angle \rm C[/tex] will be always greater than 90 degrees, that is why we say that the [tex]\rm Exterior\angle \rm C[/tex] is always obtuse.
Thus, the correct option is 1. Obtuse.
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