In an isosceles triangle, the base angles are equal. The vertex angle is $40^{o}$ . What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is $180^{o}$ ).

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Solution

Let the base angles be $^{'} x^{'}$
Therefore,
$x + x + 40^{\circ} = 180^{\circ}$ [Given]
$2 x = 180^{\circ} - 40^{\circ}$
$2 x = 140^{\circ}$
$x = 70^{\circ}$

Hence,
The base angles of the triangle are $70^{\circ}, 70^{\circ} .$

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In an isosceles triangle, the base angles are equal. The vertex angle is 40o. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180o).

Let the base angles be ′x′Therefore, x+x+40∘=180∘ [Given]2x=180∘−40∘2x=140∘x=70∘Hence, The base angles of the triangle are 70∘,70∘.

In an isosceles triangle, the base angles are equal. The vertex angle is $40^{o}$ . What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is $180^{o}$ ).

Let the base angles be $^{'} x^{'}$
Therefore,
$x + x + 40^{\circ} = 180^{\circ}$ [Given]
$2 x = 180^{\circ} - 40^{\circ}$
$2 x = 140^{\circ}$
$x = 70^{\circ}$

Hence,
The base angles of the triangle are $70^{\circ}, 70^{\circ} .$