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Identify the properties of triangle POS.

Mathematics

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3 parts
a

∠PSO=25∘\angle PSO = 25^\circ

b

∠PST=65∘\angle PST = 65^\circ

c

∠PRQ=40∘\angle PRQ = 40^\circ

Step-by-step solution

7 steps

Part a· 2 steps

Find ∠PSO\angle PSO.

  1. Identify the properties of triangle POS.

    OP and OS are both radii of the circle, so OP=OSOP = OS. This means triangle POS is an isosceles triangle.

  2. Use the property of isosceles triangles.

    In an isosceles triangle, the angles opposite the equal sides are equal. Therefore, ∠PSO=∠OPS\angle PSO = \angle OPS.
    From the diagram, we are given ∠OPS=25∘\angle OPS = 25^\circ (assuming the typo ∠OPC\angle OPC refers to ∠OPS\angle OPS).

    ∠PSO=25∘\angle PSO = 25^\circ ∠PSO=25∘\boxed{\angle PSO = 25^\circ}

Part b

∠PST=65∘\angle PST = 65^\circ

Find ∠PST\angle PST.

  1. Use the property of a tangent and radius.

    QRST is a tangent to the circle at point S. OS is the radius to the point of tangency. The radius to the point of tangency is perpendicular to the tangent.
    Therefore, ∠OST=90∘\angle OST = 90^\circ.

  2. Calculate ∠PST\angle PST.

    We know ∠OST=90∘\angle OST = 90^\circ and from part (a), ∠PSO=25∘\angle PSO = 25^\circ.

    ∠PST=∠OST−∠PSO\angle PST = \angle OST - \angle PSO ∠PST=90∘−25∘\angle PST = 90^\circ - 25^\circ ∠PST=65∘\angle PST = 65^\circ ∠PST=65∘\boxed{\angle PST = 65^\circ}

Part c

∠PRQ=40∘\angle PRQ = 40^\circ

Find ∠PRQ\angle PRQ.

  1. Find ∠POS\angle POS.

    In triangle POS, the sum of angles is 180∘180^\circ. We know ∠OPS=25∘\angle OPS = 25^\circ and ∠PSO=25∘\angle PSO = 25^\circ.

    ∠POS=180∘−(∠OPS+∠PSO)\angle POS = 180^\circ - (\angle OPS + \angle PSO) ∠POS=180∘−(25∘+25∘)\angle POS = 180^\circ - (25^\circ + 25^\circ) ∠POS=180∘−50∘\angle POS = 180^\circ - 50^\circ ∠POS=130∘\angle POS = 130^\circ

  2. Find ∠ROS\angle ROS.

    POR is a straight line, so ∠POS\angle POS and ∠ROS\angle ROS are angles on a straight line.

    ∠ROS=180∘−∠POS\angle ROS = 180^\circ - \angle POS ∠ROS=180∘−130∘\angle ROS = 180^\circ - 130^\circ ∠ROS=50∘\angle ROS = 50^\circ

  3. Find ∠PRQ\angle PRQ (which is ∠ORS\angle ORS).

    Consider triangle ORS. We know ∠OSR=90∘\angle OSR = 90^\circ (radius OS perpendicular to tangent QRST) and ∠ROS=50∘\angle ROS = 50^\circ.
    The sum of angles in triangle ORS is 180∘180^\circ.

    ∠ORS+∠OSR+∠ROS=180∘\angle ORS + \angle OSR + \angle ROS = 180^\circ ∠ORS+90∘+50∘=180∘\angle ORS + 90^\circ + 50^\circ = 180^\circ ∠ORS+140∘=180∘\angle ORS + 140^\circ = 180^\circ ∠ORS=180∘−140∘\angle ORS = 180^\circ - 140^\circ ∠ORS=40∘\angle ORS = 40^\circ

    Since ∠PRQ\angle PRQ is the same as ∠ORS\angle ORS:

    ∠PRQ=40∘\angle PRQ = 40^\circ ∠PRQ=40∘\boxed{\angle PRQ = 40^\circ}
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