Question
In an isosceles triangle, the base angles are equal. The vertex angle is $40°$. what are the base angles of the triangle? (Remember, the sum of three angles of a triangle is $180°$)
✓
Answer
Let ' $x$ ' be each base angle of the isosceles triangle
According to the question, we get
$2 x^{\circ}+40^{\circ}=180^{\circ}$
$\therefore 2 x^{\circ}=180^{\circ}-40^{\circ}$
$\therefore 2 x^{\circ}=140^{\circ}$
$\therefore x^{\circ}=\frac{140^{\circ}}{2}=70^{\circ}$
Therefore, the base angles of the triangle are $70^{\circ}$ each.
According to the question, we get
$2 x^{\circ}+40^{\circ}=180^{\circ}$
$\therefore 2 x^{\circ}=180^{\circ}-40^{\circ}$
$\therefore 2 x^{\circ}=140^{\circ}$
$\therefore x^{\circ}=\frac{140^{\circ}}{2}=70^{\circ}$
Therefore, the base angles of the triangle are $70^{\circ}$ each.
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