ABCD is a cyclic quadrilateral. $\mathrm{AB}$ and $\mathrm{DC}$ meet at $\mathrm{F}$, when produced. $\mathrm{AD}$ and $\mathrm{BC}$ meet at $\mathrm{E}$, when produced. If $\angle \mathrm{BAD}$ $=68^{\circ}$ and $\angle \mathrm{AEB}=27^{\circ}$, then what is the measure of $\angle \mathrm{BFC}$ ?
Answer & explanation
Correct answer: option 3
\(\angle\)BAE + \(\angle\)BEA + \(\angle\)ABE = 180
= 68 + 27 + \(\angle\)ABE = 180
= 95 + \(\angle\)ABE = 180
= \(\angle\)ABE = (180 - 95)
= \(\angle\)ABE = \({85}^\circ\)
Now,
\(\angle\)ABE + \(\angle\)CBF = 180 (linear pair)
= 85 + \(\angle\)CBF = 180
= \(\angle\)CBF = (180 - 85)
= \(\angle\)CBF = 95
Now,
\(\angle\)BAE + \(\angle\)BCD = 180
= 68 + \(\angle\)BCD = 180
= \(\angle\)BCD = (180 - 68)
= \(\angle\)BCD = \({112}^\circ\)
Now,
\(\angle\)BCD + \(\angle\)BCF = 180
= 112 + \(\angle\)BCF = 180
= \(\angle\)BCF = (180 - 112)
= \(\angle\)BCF = \({68}^\circ\)
Now,
In \(\Delta \)BCF
\(\angle\)BCF + \(\angle\)CBF + \(\angle\)BFC = 180
= 68 + 95 +\(\angle\)BFC = 180
= 163 + \(\angle\)BFC = 180
= \(\angle\)BFC = (180 - 163)
= \(\angle\)BFC = \({17}^\circ\)
Therefore, answer is \({17}^\circ\).