Math

QuestionFind angle OQPO Q P given points P,Q,RP, Q, R on a circle with mQPR=35m\angle Q P R=35^{\circ} and mORP=30m\angle O R P=30^{\circ}.

Studdy Solution

STEP 1

Assumptions1. , $Q$, and $R$ are points on a circle with center . . Angle QPR=35QPR =35^{\circ} and angle OR=30OR =30^{\circ}.
3. Line segments QQ and PRPR intersect at point XX.
4. We need to find angle QPQP.

STEP 2

We know that the angles in a triangle add up to 180180^{\circ}. So, in triangle OROR, we can find angle OPROPR by subtracting the given angles from 180180^{\circ}.
OPR=180ORQPR\angle OPR =180^{\circ} - \angle OR - \angle QPR

STEP 3

Now, plug in the given values for the angles to calculate angle OPROPR.
OPR=1803035\angle OPR =180^{\circ} -30^{\circ} -35^{\circ}

STEP 4

Calculate the angle OPROPR.
OPR=1803035=115\angle OPR =180^{\circ} -30^{\circ} -35^{\circ} =115^{\circ}

STEP 5

We know that the angle at the center is twice the angle at the circumference when the two angles subtend the same arc. Therefore, angle QPQP is half of angle OPROPR.
OQP=12OPR\angle OQP = \frac{1}{2} \angle OPR

STEP 6

Now, plug in the value for angle OPROPR to calculate angle QPQP.
OQP=12×115\angle OQP = \frac{1}{2} \times115^{\circ}

STEP 7

Calculate the angle QPQP.
OQP=12×115=57.5\angle OQP = \frac{1}{2} \times115^{\circ} =57.5^{\circ}The angle QPQP is 57.557.5^{\circ}.

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