Question

one of the exterior angles of an isosceles triangle is 115 degrees. find the angles of this triangle.

Answer 1

In an isosceles triangle, two sides are equal and the angles opposite those sides are also equal. We know one of the exterior angles of the triangle is 115 degrees.

An exterior angle of a triangle is equal to the sum of the two opposite interior angles. Let's denote the interior angles of the isosceles triangle as \( A \), \( A \) (the two equal angles), and \( B \) (the vertex angle). Since the exterior angle is given as 115 degrees, we have:

\[
A + B = 115^\circ
\]

We also know that the sum of all interior angles in a triangle is \( 180^\circ \):

\[
A + A + B = 180^\circ
\]
\[
2A + B = 180^\circ
\]

Now we have a system of two equations:

1. \( A + B = 115^\circ \)
2. \( 2A + B = 180^\circ \)

We can solve these equations step by step. First, we can isolate \( B \) from the first equation:

\[
B = 115^\circ - A
\]

Now, we can substitute this expression for \( B \) into the second equation:

\[
2A + (115^\circ - A) = 180^\circ
\]

This simplifies to:

\[
2A - A + 115^\circ = 180^\circ
\]
\[
A + 115^\circ = 180^\circ
\]
\[
A = 180^\circ - 115^\circ
\]
\[
A = 65^\circ
\]

Now that we have \( A \), we can find \( B \) using the first equation:

\[
B = 115^\circ - A = 115^\circ - 65^\circ = 50^\circ
\]

Thus, the angles of the isosceles triangle are:

\[
A = 65^\circ, \quad A = 65^\circ, \quad B = 50^\circ
\]

So, the angles of the triangle are \( 65^\circ, 65^\circ, \) and \( 50^\circ \).