The vertex angle of an isosceles triangle is 10 less than three times the base angles. Find each angle of the triangle.

From the picture attached,

       ∠B ≅ ∠C

       m(∠A) = 3[m(∠B)] - 10

By applying angle sum theorem in ΔABC,

m(∠A) + m(∠B) + m(∠C) = 180°

m(∠A) + m(∠B) + m(∠B) = 180° [From equation (1)]

m(∠A) + 2m(∠B) = 180°

[3m(∠b) - 10] + 2m(∠B) = 180°

5[m(∠B)] = 180° + 10°

m∠B = 38°

       Therefore, m∠B = m∠C = 38°

Substitute the value of angle B in equation (2),

m(∠A) = 3(38) - 10

          = 104°

        Therefore, measure of angle A will be 104°.

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