Line CP parallel to line D is drawn through vertex C of triangle CDE with right angle D.

Line CP parallel to line D is drawn through vertex C of triangle CDE with right angle D. Find angles C and E of triangle CDE if angle PCE = 49 degrees.

Since, according to the condition, the line CP is parallel to the line DE, then the angle CED = BCE = 49, as criss-crossing angles at the intersection of parallel lines CP and BD secant CE.

Then, in a right-angled triangle CDE, the angle is DCE = 180 – CDE – CED = 180 – 90 – 49 = 41.

Answer: Angle C = 41, Angle E = 49.