In an isosceles triangle is one of the angles 40∘. What values are possible for the outer angles.
(not asking help from Writeacher) only oobleck
The sum of the measures of the angles is 180° in a triangle.
An isosceles triangle has 2 equal sides and 2 equal angles.
Mark equal angles with θ.
So:
40° + θ + θ = 180°
40° + 2 θ = 180°
Subtract 40° to both sides
2 θ = 140°
Divide both sides by 2
θ = 70°
In this triangle the angles are:
40° , 70° , 70°
The other possibility is, of course,
40°, 40°, 100°
To find the possible values of the outer angles in an isosceles triangle, we can first determine the measure of the base angles.
In an isosceles triangle, the base angles are equal. Let's assume the measure of the base angles is x.
Since we're given that one of the angles is 40°, and the base angles are equal, we can set up an equation:
40° + x + x = 180° (the sum of angles in a triangle is 180°)
Simplifying the equation:
40° + 2x = 180°
Now, let's solve for x by isolating it on one side:
2x = 180° - 40°
2x = 140°
x = 140° / 2
x = 70°
Therefore, the base angles of the isosceles triangle are each 70°.
Now, let's find the values of the outer angles. To do this, we subtract each base angle from 180°, as the sum of an outer angle and its corresponding base angle is 180°.
First outer angle: 180° - 70° = 110°
Second outer angle: 180° - 70° = 110°
So, the possible values for the outer angles in the isosceles triangle are both 110°.