Question

in parallelogram (abcd), (mangle aeb = 112^circ), (mangle cdb = 27^circ), (mangle adb = 58^circ), (bc = 6), (be = 7), and (ce = 5). find the measure of (angle bac), (angle bec), and (angle dac).
note: the diagram is not drawn to scale.
(mangle bac =)
(mangle bec =)
(mangle dac =)

Explanation:

Step1: Find \( m\angle BAC \)

In parallelogram \(ABCD\), \(AB \parallel CD\), so \( \angle BAC=\angle DCA \) (alternate interior angles). Also, in \( \triangle CDB \), we know \( \angle CDB = 27^{\circ}\) and \( \angle ADB = 58^{\circ}\), so \( \angle ADC=\angle ADB+\angle CDB = 58^{\circ}+27^{\circ}=85^{\circ}\). In a parallelogram, \( \angle ADC+\angle BCD = 180^{\circ}\), but we can also use the fact that \( \angle BAC\) and \( \angle CDB\) are related? Wait, no. Wait, \(AB \parallel CD\), so \( \angle ABD=\angle CDB = 27^{\circ}\) (alternate interior angles). In \( \triangle AEB\), we know \( \angle AEB = 112^{\circ}\), and \( \angle ABD = 27^{\circ}\), so \( \angle BAC=180^{\circ}-\angle AEB-\angle ABD=180^{\circ}-112^{\circ}-27^{\circ}=41^{\circ}\)? Wait, no, \( \angle ABD\) is \( \angle ABE\), which is equal to \( \angle CDB = 27^{\circ}\) because \(AB \parallel CD\) and \(BD\) is a transversal. So in \( \triangle AEB\), angles sum to \(180^{\circ}\), so \( \angle BAC = 180^{\circ}-112^{\circ}-27^{\circ}=41^{\circ}\)? Wait, no, \( \angle ABE\) is \( \angle ABD\), which is equal to \( \angle CDB = 27^{\circ}\) (alternate interior angles). So \( \angle BAC = 180 - 112 - 27 = 41^{\circ}\).

Step2: Find \( m\angle BEC \)

\( \angle AEB\) and \( \angle BEC\) are supplementary because they form a linear pair. So \( \angle BEC = 180^{\circ}-\angle AEB = 180^{\circ}-112^{\circ}=68^{\circ}\).

Step3: Find \( m\angle DAC \)

In parallelogram \(ABCD\), \(AD \parallel BC\), so \( \angle DAC=\angle BCA \) (alternate interior angles). Also, in \( \triangle ADB\), \( \angle ADB = 58^{\circ}\), and \(AD = BC = 6\) (opposite sides of parallelogram). Wait, \( \angle DAC\): let's look at \( \triangle ADE\) or \( \triangle ADC\). Wait, \( \angle ADB = 58^{\circ}\), and \(AB \parallel CD\), so \( \angle BAD + \angle ADC = 180^{\circ}\), \( \angle ADC = 85^{\circ}\), so \( \angle BAD = 95^{\circ}\). \( \angle BAD=\angle BAC+\angle DAC\), so \( \angle DAC=\angle BAD - \angle BAC\). We found \( \angle BAC = 41^{\circ}\), \( \angle BAD = 180^{\circ}-\angle ADC = 180^{\circ}-85^{\circ}=95^{\circ}\), so \( \angle DAC = 95^{\circ}-41^{\circ}=54^{\circ}\)? Wait, no, \( \angle ADC = \angle ADB + \angle CDB = 58 + 27 = 85^{\circ}\), so \( \angle BAD = 180 - 85 = 95^{\circ}\). \( \angle BAD = \angle BAC + \angle DAC\), so \( \angle DAC = 95 - 41 = 54^{\circ}\). Alternatively, in \( \triangle ADB\), \( \angle DAC\) is equal to \( \angle BCA\), and since \(BC = 6\), \(AD = 6\), and \(BE = 7\), \(CE = 5\), but maybe easier: \( \angle DAC\): \( \angle ADB = 58^{\circ}\), and \( \angle DAC\) and \( \angle ADB\) in \( \triangle ADE\)? Wait, no. Wait, \( \angle DAC\): let's use the fact that \( \angle ADB = 58^{\circ}\), and \(AB \parallel CD\), so \( \angle ABD = 27^{\circ}\), \( \angle BAD = 95^{\circ}\), so \( \angle DAC = \angle BAD - \angle BAC = 95 - 41 = 54^{\circ}\). Wait, but let's check: \( \angle ADB = 58^{\circ}\), and \( \angle DAC\): in \( \triangle ADC\), \( \angle ADC = 85^{\circ}\), \( \angle DCA = \angle BAC = 41^{\circ}\), so \( \angle DAC = 180 - 85 - 41 = 54^{\circ}\). Yes, that works.

Answer:

\( m\angle BAC = 41^{\circ} \)
\( m\angle BEC = 68^{\circ} \)
\( m\angle DAC = 54^{\circ} \)