Question

in the diagram, ( ad = cd = cb ) and ( mangle a = 40^circ ). how many degrees are in ( angle dcb )?

Explanation:

Step1: Analyze triangle \( \triangle ADC \)

Since \( AD = CD \), \( \triangle ADC \) is isosceles with \( \angle A=\angle ACD = 40^{\circ} \). The sum of angles in a triangle is \( 180^{\circ} \), so \( \angle ADC=180^{\circ}-40^{\circ}-40^{\circ}=100^{\circ} \).

Step2: Find \( \angle CDB \)

\( \angle ADC \) and \( \angle CDB \) are supplementary (they form a linear pair), so \( \angle CDB = 180^{\circ}-\angle ADC=180^{\circ}-100^{\circ}=80^{\circ} \).

Step3: Analyze triangle \( \triangle CDB \)

Since \( CD = CB \), \( \triangle CDB \) is isosceles with \( \angle CDB=\angle B = 80^{\circ} \). The sum of angles in a triangle is \( 180^{\circ} \), so \( \angle DCB=180^{\circ}-\angle CDB-\angle B=180^{\circ}-80^{\circ}-80^{\circ}=20^{\circ} \).

Answer:

20