In Fig. ∠X = 62°, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of Δ XYZ, find ∠O.

Solution:

As we know, the sum of the interior angles of the triangle is 180°.

So, ∠X +∠XYZ + ∠XZY = 180°

substituting the values as given in the question, we get:

62° + 54° + ∠XZY = 180°

Or, ∠XZY = 64°

Now, As we know, ZO is the bisector so,

∠OZY = $\frac{1}{2}$∠XZY

∴ ∠OZY = 32°

Similarly, YO is a bisector and so,

∠OYZ = $\frac{1}{2}$∠XYZ

Or, ∠OYZ = 27° (As ∠XYZ = 54°)

Now, as the sum of the interior angles of the triangle,

∠OZY +∠OYZ + ∠O = 180°

Substituting their respective values we get,

∠O = 180° – 32° – 27°

Or, ∠O = 121°