Question

In rectangle $ABCD$ below, we have $DP = PC$ and $\angle BAC = 28^\circ$. If $Q$ is the point on $\overline{AC}$ such that $\overline{DQ}\perp \overline{AC}$, what is $\angle QDP$ in degrees?

Answer 1

∠QDP = 34°

Explanation:

The complete question is shown in the image attached.

In triangle ABC, ∠ABC = 90° and ∠BCA = 28°. Hence:

∠CAB + ∠ABC + ∠BCA = 180° (sum of angles in a triangle)

∠BCA + 28+ 90 = 180

∠BCA + 118 = 180

∠BCA = 180 - 118

∠BCA = 62°

Also ∠DCA + ∠BCA = 90° (Each angle in a rectangle is 90°)

∠DCA + 62 = 90

∠DCA = 90 - 62 = 28°

∠DAC + ∠BAC = 90 (angle in a rectangle)

∠DAC + 28 = 90

∠DAC = 90 - 28 = 62°

In triangle ADQ, ∠DQA = 90° and ∠QAD = 62°. Hence:

∠QAD + ∠ADQ + ∠QAD = 180° (sum of angles in a triangle)

∠ADQ + 62+ 90 = 180

∠ADQ + 152 = 180

∠ADQ = 180 - 152

∠ADQ = 28°

In triangle DPC,DP = PC, hence ∠PDC = ∠DCP= 28°.

∠PDC + ∠ADQ + ∠QDP = 90° (angle in a rectangle)

28 + 28 + ∠QDP = 90

∠QDP + 56 = 90

∠QDP = 90 - 56

∠QDP = 34°