ABCD is a cyclic quadrilateral. Diagonals BD and AC intersect each other at E. If ∠BEC = 138° and ∠ECD = 35°, then what is the measure of ∠BAC ?
Answer & explanation
Correct answer: option 4
\(\angle\)BEC and \(\angle\)CED are on the same straight lines
\(\angle\)BEC = 138
\(\angle\)CED = 180 - 138
= \(\angle\)CED = \({42}^\circ\)
In \(\Delta \)CDE, \(\angle\)CED = 42 and \(\angle\)DCE = 35
\(\angle\)CDE = 180 - (42 + 35)
= \(\angle\)CDE = \({103}^\circ\)
\(\angle\)BAC and \(\angle\)BDC are on the same arc BC
We know that in cyclic quadrilateral angles on the same arc are always same.
\(\angle\)BAC = \({103}^\circ\)
Therefore, \(\angle\)BAC is \({103}^\circ\).