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

Given: ∠X = 62^o, ∠ XYZ = 54^o

YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively.
To Find ∠ OZY and ∠ YOZ.

Now, according to the question,

∠X + ∠XYZ + ∠XZY = 180^o(Sum of the interior angles of a triangle = 180°)

62^o + 54^o + ∠XZY = 180^o

116^o + ∠XZY = 180^o

∠XZY = 64^o

Now,

As ZO is the bisector of ∠ XZY

∠OZY = 1/2 ∠ XZY

∠OZY = 32^o

And,

As YO is bisector of ∠ XYZ

∠OYZ = 1/2 ∠XYZ

∠OYZ = 27^o

Now,

∠OZY + ∠OYZ + ∠O = 180^o(Sum of the interior angles of the triangle = 180°)

32^o + 27^o + ∠O = 180^o

59^o + ∠O = 180^o

∠O = 121^o