An interior angle of an isosceles triangle is $p^0$ and an exterior angle is $q^0$ - OCR - GCSE Maths - Question 10 - 2019 - Paper 5

To find the different possible sets of angles for the isosceles triangle, we can use the relationships between the interior and exterior angles.

1. The exterior angle $q$ is equal to the sum of the two interior angles. Since it's an isosceles triangle:
   • Let the two equal interior angles be $p$.
   • The relation is:
     $q = 2p$
2. We already have $q = 5p$, so we can set up the equation:
   $5p = 2p$
   This leads to:
   $5p - 2p = 0$
   $3p = 0$
   Therefore, $p = 0$, which is not a valid angle.
3. We need to acknowledge that the angle $p$ must also satisfy the triangle sum property. Thus:
   • Total sum of angles in a triangle = $180^0$.
   • Therefore, $2p + q = 180^0$ implies:
     $2p + 5p = 180$
   • Simplifying gives:
     $7p = 180$
     p = rac{180}{7} ext{ degrees}
4. Thus, we have one angle set:
   • p = rac{180}{7}^0, q = 5p = rac{900}{7}^0.
5. For the second possibility, consider another configuration which still adheres to the properties of isosceles triangles leading to a similar calculation yet yielding valid angles $60^0$ and $60^0$.

In conclusion, the two sets of angles for the isosceles triangle possible are:

1. p = rac{180}{7}^0, q = rac{900}{7}^0
2. $p = 60^0$, and each angle must add up to $180^0$.