1) One angle of an isosceles triangle measures $$ 140^{\circ} $$. What measures are possible for the other two angles? Choose all that apply.

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In an isosceles triangle, two angles are equal. Let's denote the angles of the triangle as $$ A $$, $$ B $$, and $$ C $$. Given that one angle measures $$ 140^{\circ} $$, we can consider two cases:

1. **Case 1:** The angle measuring $$ 140^{\circ} $$ is one of the equal angles.
2. **Case 2:** The angle measuring $$ 140^{\circ} $$ is the unique angle.

### Case 1: $$ A = B = 140^{\circ} $$

If both equal angles are $$ 140^{\circ} $$, then the third angle $$ C $$ can be calculated using the triangle sum property, which states that the sum of the angles in a triangle is $$ 180^{\circ} $$:

$$
A + B + C = 180^{\circ}
$$

Substituting the values:

$$
140^{\circ} + 140^{\circ} + C = 180^{\circ}
$$

This simplifies to:

$$
280^{\circ} + C = 180^{\circ}
$$

Solving for $$ C $$:

$$
C = 180^{\circ} - 280^{\circ} = -100^{\circ}
$$

This is not possible since angles cannot be negative.

### Case 2: $$ A = 140^{\circ} $$ and $$ B = C $$

In this case, let $$ B = C = x $$. Again, using the triangle sum property:

$$
A + B + C = 180^{\circ}
$$

Substituting the values:

$$
140^{\circ} + x + x = 180^{\circ}
$$

This simplifies to:

$$
140^{\circ} + 2x = 180^{\circ}
$$

Solving for $$ x $$:

$$
2x = 180^{\circ} - 140^{\circ}
$$
$$
2x = 40^{\circ}
$$
$$
x = 20^{\circ}
$$

Thus, in this case, the other two angles are both $$ 20^{\circ} $$.

### Conclusion

The possible measures for the other two angles in the isosceles triangle are:

- $$ 20^{\circ} $$ and $$ 20^{\circ} $$ (the two equal angles).

Therefore, the only valid measures for the other two angles are $$ 20^{\circ} $$.
The other two angles are both $$ 20^{\circ} $$.

1) One angle of an isosceles triangle measures \( 140^{\circ} \). What measures are possible for the other two angles? Choose all that apply.

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