In the diagram, \(\bar{OX}\) bisects < YXZ and \(\bar{OZ}\) bisects < YZX. If < XYZ = 68o, calculate the value of < XOZ

In the diagram, \(\bar{OX}\) bisects < YXZ and \(\bar{OZ}\) bisects < YZX. If < XYZ = 68o, calculate the value of < XOZ

A.

68o

B.

72o

C.

112o

D.

124o

Correct answer is D

In \(\Delta\) XYZ, 2m + 2n + 68o = 180o

2(m + n) + 68o = 180o...(1)

in \(\Delta\) XOZ, m + n + q = 180o ...(2)

(m + n) = 180o - q...(3)

substituting 180o - q for (m + n) in (1) gives

2(180o - q) + 68o = 180o

360o - 2q = 180o - 68o

360o - 2q = 112o

360o - 112o = 2q

248o = 2q

q = \(\frac{248^o}{2}\)

= 124o

hence, < XOZ = 124o