Me and mine solution math with explanetion
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Given: The given triangle is isosceles triangle. Also vertex angle is 100°
To find: Measure of base angles.
It is given that triangle is isosceles.
So let our given triangle be ∆ABC.
And let ∠A be the vertex angle, which is given as ∠A= 100°
By the property of isosceles triangle, we know that base angles are equal.
So,
∠B = ∠C …(1)
We know that,
Sum of all angles in any triangle = 180°
∴ ∠A + ∠B + ∠C = 180°
100° + 2∠B = 180° …from (1)
∴ 2∠B = 180° - 100°
2∠B = 80°
∴∠B = 40°
Therefore, our base angles, ∠B and ∠C, are 40° each.
To find: Measure of base angles.
It is given that triangle is isosceles.
So let our given triangle be ∆ABC.
And let ∠A be the vertex angle, which is given as ∠A= 100°
By the property of isosceles triangle, we know that base angles are equal.
So,
∠B = ∠C …(1)
We know that,
Sum of all angles in any triangle = 180°
∴ ∠A + ∠B + ∠C = 180°
100° + 2∠B = 180° …from (1)
∴ 2∠B = 180° - 100°
2∠B = 80°
∴∠B = 40°
Therefore, our base angles, ∠B and ∠C, are 40° each.
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