One of the outer corners of an isosceles triangle is 110 °. find the measure of the lesser

| June 10, 2021 | education

One of the outer corners of an isosceles triangle is 110 °. find the measure of the lesser of the inner angles of this triangle.

a) The vertices of the triangle A, B, C. AB = BC. Outside apex angle B = 110 °.

1. We calculate the degree measure of the internal ∠В, taking into account that the sum of adjacent angles is equal to 180 °:

∠В = 180 ° – 110 ° = 70 °.

2. By the condition of the problem, the given triangle is isosceles. Therefore, ∠A = ∠C.

3. We calculate the degree measure of the above angles:

∠А = ∠С = (180 ° – 70 °) / 2 = 55 °.

Answer: ∠А = ∠С – smaller angles of a given triangle.

c) The vertices of the triangle A, B, C. AB = BC. The external apex angle is A = 110 °.

1. We calculate the degree measure of the internal ∠А, taking into account that the sum of adjacent angles is equal to 180 °:

∠А = 180 ° – 110 ° = 70 °.

2. ∠А = ∠С = 70 °.

3. We calculate the degree measure ∠В:

∠В = 180 ° – 70 ° – 70 ° = 40 °.

Answer: ∠В = 40 ° – the smaller angle of the given triangle.



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