Two angles of a triangle are equal and the third angle is greater than each of those angles by $30^o$. Determine all the angles of the triangle.
Given:
Two angles of a triangle are equal and the third angle is greater than each of those angles by $30^o$.
To do:
We have to determine all the angles of the triangle.
Solution:
We know that,
Sum of the angles in a triangle is $180^o$.
Let the two equal angles each be $x$.
This implies,
The third angle $=x+30^o$
Therefore,
$x+x+x+30^o=180^o$
$3x=180^o-30^o$
$3x=150^o$
$x=50^o$
$x+30^o=(50+30)^o=80^o$
Hence, the angles of the triangle are $50^o, 50^o$ and $80^o$.
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