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In the diagram, |PQ| = |PR| = |RS| and ∠RPS = 32°. Find ∠QPR.

Mathematics

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Answer

D

Step-by-step solution

4 steps

I will solve question 8.

Question 8: In the diagram, ∣PQ∣=∣PR∣=∣RS∣|PQ| = |PR| = |RS| and ∠RPS=32∘\angle RPS = 32^\circ. Find ∠QPR\angle QPR.

  1. Analyze △PRS\triangle PRS.

    Given that ∣PR∣=∣RS∣|PR| = |RS|, △PRS\triangle PRS is an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are equal.
    Therefore, ∠RSP=∠RPS\angle RSP = \angle RPS.
    Since ∠RPS=32∘\angle RPS = 32^\circ, then ∠RSP=32∘\angle RSP = 32^\circ.

  2. Find the exterior angle ∠PRQ\angle PRQ of △PRS\triangle PRS.

    The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

    ∠PRQ=∠RPS+∠RSP\angle PRQ = \angle RPS + \angle RSP ∠PRQ=32∘+32∘\angle PRQ = 32^\circ + 32^\circ ∠PRQ=64∘\angle PRQ = 64^\circ
  3. Analyze △PQR\triangle PQR.

    Given that ∣PQ∣=∣PR∣|PQ| = |PR|, △PQR\triangle PQR is an isosceles triangle.
    Therefore, ∠PQR=∠PRQ\angle PQR = \angle PRQ.
    From Step 2, we found ∠PRQ=64∘\angle PRQ = 64^\circ.
    So, ∠PQR=64∘\angle PQR = 64^\circ.

  4. Find ∠QPR\angle QPR in △PQR\triangle PQR.

    The sum of angles in a triangle is 180∘180^\circ.

    ∠QPR+∠PQR+∠PRQ=180∘\angle QPR + \angle PQR + \angle PRQ = 180^\circ ∠QPR+64∘+64∘=180∘\angle QPR + 64^\circ + 64^\circ = 180^\circ ∠QPR+128∘=180∘\angle QPR + 128^\circ = 180^\circ ∠QPR=180∘−128∘\angle QPR = 180^\circ - 128^\circ ∠QPR=52∘\angle QPR = 52^\circ

    The value of ∠QPR\angle QPR is 52∘52^\circ.

    The correct option is D.

    D. 52∘\boxed{\text{D. } 52^\circ}

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