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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Updated on: 10-Oct-2022

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