Question

58.
m∠ace =

m∠acd =

m∠bcf =

Explanation:

Step1: Find \( m\angle ACE \)

We know that \( \angle ACB \) is a straight angle? No, wait, \( \angle ACF \) and \( \angle FCE \) and \( \angle ECA \)? Wait, looking at the diagram, \( \angle ACF \) is a right angle? Wait, the square at \( \angle ACB \)? Wait, no, the diagram has \( \angle ACF \) and \( \angle FCE = 12^\circ \), and \( \angle ACF \) is a right angle? Wait, \( \angle ACE + \angle ECF + \angle FCB \)? Wait, no, let's re - examine. The angle \( \angle ACB \) is a straight line? No, the square at \( C \) for \( \angle ACF \) and \( \angle FCB \)? Wait, actually, \( \angle ACF \) is a right angle? Wait, the diagram shows that \( \angle ACF \) is a right angle (the square symbol), so \( \angle ACF=90^\circ \). And \( \angle ECF = 12^\circ \), so \( \angle ACE=\angle ACF-\angle ECF \).
\( m\angle ACE = 90^\circ- 12^\circ=78^\circ \)

Step2: Find \( m\angle ACD \)

We know that \( \angle BCD = 105^\circ \), and \( \angle ACB \) is a straight angle (\( 180^\circ \)). So \( \angle ACD=180^\circ - \angle BCD \)
\( m\angle ACD=180^\circ - 105^\circ = 75^\circ \)

Step3: Find \( m\angle BCF \)

From the diagram, \( \angle BCF \) is a right angle (the square symbol), so \( m\angle BCF = 90^\circ \)

Answer:

\( m\angle ACE=\boldsymbol{78^\circ} \), \( m\angle ACD=\boldsymbol{75^\circ} \), \( m\angle BCF=\boldsymbol{90^\circ} \)