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