Answer: The two possible cases are
(i) the base angles are each 40° and the vertex angle is 100°,
(ii) the base angles are each 70° and the vertex angle is 40°.
Step-by-step explanation: Given that an exterior angle of an isosceles triangle has measure 140 degrees.
We are to find the two possible sets of measures for the angles of the triangle.
As shown in the attached figure, ΔABC is an isosceles triangle, where AB = AC.
So, ∠ABC = ∠ACB.
First case : Let the exterior angle of the bases is 140 degrees°.
That is, ∠ABE = ∠ACF = 140°.
From the property of a linear pair, we have
[tex]\angle ABE+\angle ABC=180^\circ\\\\\Rightarrow 140^\circ+\angle ABC=180^\circ\\\\\Rightarrow \angle ABC=180^\circ-140^\circ\\\\\Rightarrow \angle ABC=40^\circ.[/tex]
So, ∠ABC = ∠ACB = 40°.
We know that the sum of three angles of a triangle is 180°, so
[tex]\angle ABC+\angle ACB+\angle BAC=180^\circ\\\\\Rightarrow 40^\circ+40^\circ+\angle BAC=180^\circ\\\\\Rightarrow \angle BAC=100^\circ.[/tex]
Thus, the base angles are each 40° and the vertex angle is 100°.
Second case: Let the exterior angle of the vertex is 140 degrees.
That is, ∠CAD = 140°.
From the property of a linear pair, we have
[tex]\angle CAD+\angle BAC=180^\circ\\\\\Rightarrow 140^\circ+\angle BAC=180^\circ\\\\\Rightarrow \angle BAC=180^\circ-140^\circ\\\\\Rightarrow \angle BAC=40^\circ.[/tex]
We know that the sum of three angles of a triangle is 180°, so
[tex]\angle ABC+\angle ACB+\angle BAC=180^\circ\\\\\Rightarrow 2\angle ABC+40^\circ=180^\circ\\\\\Rightarrow 2\angle ABC=180^\circ-40^\circ\\\\\Rightarrow 2\angle ABC=140^\circ\\\\\Rightarrio \angle ABC=70^\circ.[/tex]
So, ∠ABC = ∠ACB = 70° and ∠BAC = 40°.
Thus, the base angles are each 70° and the vertex angle is 40°.
Hence, the two possible cases are
(i) the base angles are each 40° and the vertex angle is 100°,
(ii) the base angles are each 70° and the vertex angle is 40°.
