Answer:
[tex]55^{\circ}[/tex] and [tex]55^{\circ}[/tex].
Step-by-step explanation:
Refer to the diagram attached. In an isosceles triangle ([tex]\triangle {\sf ABC}[/tex],) two of the sides are of equal length (the legs, [tex]{\sf AB}[/tex] and [tex]{\sf AC}[/tex] in the diagram.) The two angles adjacent to the third side (the base, [tex]{\sf BC}[/tex]) would be equal: [tex]\angle {\sf B} = \angle {\sf C}[/tex].
In this question, it is given that only one of the three angles measured [tex]70^{\circ}[/tex]. As the two angles adjacent to the base ([tex]\angle {\sf B}[/tex] and [tex]\angle {\sf C}[/tex]) are of equal measure, this [tex]70^{\circ}[/tex] angle cannot be either of them. Rather, that angle must be opposite to the base: [tex]\angle {\sf A} = 70^{\circ}[/tex].
The sum of the three angles in a triangle should be [tex]180^{\circ}[/tex]. In other words:
[tex]\angle {\sf A} + \angle {\sf B} + \angle {\sf C} = 180^{\circ}[/tex].
Since [tex]\angle {\sf A} = 70^{\circ}[/tex] while [tex]\angle {\sf B} = \angle {\sf C}[/tex]:
[tex]\angle {\sf B} + \angle {\sf C} = 180^{\circ} - 70^{\circ} = 110^{\circ}[/tex].
[tex]\displaystyle \angle {\sf B} = \frac{\angle {\sf B} + \angle {\sf B}}{2} = \frac{\angle {\sf B} + \angle {\sf C}}{2} = \frac{110^{\circ}}{2} = 55^{\circ}[/tex].
[tex]\angle {\sf C} = \angle {\sf B} = 55^{\circ}[/tex].
Hence, the measure of the two other angles would both be [tex]55^{\circ}[/tex].
