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ivanzaharov [21]
3 years ago
15

Quadrilateral OPQR is inscribed in circle N, as shown below. Which of the following could be used to calculate the measure of ∠P

QR? Circle N is shown with a quadrilateral OPQR inscribed inside it. Angle O is labeled x plus 16. Angle P is not labeled. Angle Q is labeled 6x minus 4. Angle R is labeled 2x plus 16.

Mathematics
1 answer:
Rama09 [41]3 years ago
4 0

Answer:

m \angle PQR=80^{\circ}

Step-by-step explanation:

Missing figure drawn.

The measure of ∠PQR is found by the property of inscribed quadrilateral. In an inscribed quadrilateral the opposite angles are supplementary:

\left\{\begin{matrix}\angle O+\angle Q=180^{\circ} & \\ \angle P +\angle R=180^{\circ} & \end{matrix}\right.

x+16+6x-4=180^{\circ}\\7x+12=180\\7x+12-12=180-12\\7x=168\\\frac{7x}{7}=\frac{168}{7}\\x=14\\\angle Q=6(14)-4\Rightarrow m\angle Q=84-4\Rightarrow Q=80^{\circ}

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(a) <TOR=pi/3 radians

To determine <TOR we use the fact that in the right-angled triangle ORT we know two sides:

|OT|=radius=8cm and |OR|=radius/2=4cm

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\sin \angle OTR=\frac{r/2}{r}=\frac{1}{2}\implies \angle OTR =\frac{\pi}{6}

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(b) The arc length is approximately 7.255 cm

In order to calculate the arc length QT, we need to first determine the length |ST| and the angle <OST.

Towards determining angle <OST:

\angle SOT = \pi - \angle TOR = \pi - \frac{\pi}{3} = \frac{2}{3}\pi

Next, draw a line connecting P and T. Realize that triangle PTS is right-angled with <PTS=pi/2. This follows from the Thales theorem. Since R is a midpoint between P and O, it follows that the triangles ORT and PRT are congruent. So the angles <PTR and <OTR are congruent. Knowing <PTS we can  determine angle <OTS:

\angle OTR \cong \angle PTR=\frac{\pi}{6}\implies\angle OTS=\angle PTS -\angle PTR -\angle OTR\\\angle OTS = \frac{\pi}{2}-\frac{\pi}{6}-\frac{\pi}{6}=\frac{\pi}{6}

and so the angle <OST is

\angle OST = \pi - \angle TOS - \angle OTS = \pi -\frac{2}{3}\pi - \frac{1}{6}\pi=\frac{\pi}{6}

Towards determining |TS|:

Use cosine:

\cos \angle OST =\frac{|RS|}{|ST|}\implies |ST|=\frac{\frac{3}{2}r}{\cos \frac{\pi}{6}}=\frac{12\cdot 2}{\sqrt{3}}=8\sqrt{3}cm

Finally, we can determine the arc length QT:

QT = {\angle OST}\cdot |ST|=\frac{\pi}{6}\cdot 8 \sqrt{3}=\frac{4\pi}{\sqrt{3}}\approx 7.255cm




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