Asked by phani yej
In an isoceles triangle top vertex is twice the of the sum of the base angle.find all the exterior angles of the triangle?
Answers
Answered by
Bosnian
If it is an isosceles triangle the two base angles will be equal.
Let the base angle be a, then the vertex angle will be:
180° - 2 a
According to the condition:
Top vertex is twice the of the sum of the base angle:
180° - 2 a = 2 ( a + a )
180° - 2 a = 2 * 2a
180° - 2 a = 4 a Add 2 a to both sides
180° - 2 a + 2 a = 4 a + 2 a
180° = 6 a Divide both sides by 6
180° / 6 = a
30° = a
a = 30 °
So the base angles are 30° and the vertex angle is 180° - 2 * 30° = 180° - 60° = 120 °
Exterior base angle is 360° - 30° = 330°
The exterior vertex angle is 360° - 120° = 240°
Let the base angle be a, then the vertex angle will be:
180° - 2 a
According to the condition:
Top vertex is twice the of the sum of the base angle:
180° - 2 a = 2 ( a + a )
180° - 2 a = 2 * 2a
180° - 2 a = 4 a Add 2 a to both sides
180° - 2 a + 2 a = 4 a + 2 a
180° = 6 a Divide both sides by 6
180° / 6 = a
30° = a
a = 30 °
So the base angles are 30° and the vertex angle is 180° - 2 * 30° = 180° - 60° = 120 °
Exterior base angle is 360° - 30° = 330°
The exterior vertex angle is 360° - 120° = 240°
Answered by
Steve
I believe exterior angles are 180 - interior angles. Their sum is 360.
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