question_answer

In the given figure, EF || GH, \[\angle \]EAB = \[60{}^\circ \] and \[\angle \]ACH =\[105{}^\circ \], then find the values of:

(a) \[\angle \]CAF,

(b) \[\angle \]BAC

 

Answer:

(a) Since EF||GH and AC is transversal.

\[\angle CAF+\angle ACH=180{}^\circ \]

(interior angles on same side of transversal)

\[\angle CAF+105{}^\circ =180{}^\circ \]

\[\angle CAF=180{}^\circ 105{}^\circ \]

= 75°

(b) Since EAF is a straight line.

∴ \[\angle EAB+\angle BAC+\angle CAF=180{}^\circ \]

\[60{}^\circ +\angle BAC+75{}^\circ =180{}^\circ \]

\[\angle BAC=180{}^\circ 135{}^\circ \]

\[=45{}^\circ \]