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Zigmanuir [339]
3 years ago
8

Quadrilateral OPQR is inscribed inside a circle as shown below. Write a proof showing that angles O and Q are supplementary

Mathematics
1 answer:
k0ka [10]3 years ago
4 0
Hello,

very simple they intercept supplementary arcs. (making a sum of 360°/2 for inscribed angles )
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The point(3,7) is reflected in the y axis what are the coordinates now
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When a point P(a, b) is reflected about the y-axis, the coordinates of the reflected point are P'(-a, b).

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BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

\large \boxed{sin \theta = tan \theta cos \theta} \\  \large \boxed{tan^{2}  \theta + 1 =  {sec}^{2} \theta}

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

\large{ {4}^{2}  + 1 =  {sec}^{2} \theta } \\  \large{16 + 1 =  {sec}^{2} \theta } \\  \large{ {sec}^{2}  \theta = 17}

As we know, sec²θ = 1/cos²θ.

\large \boxed{sec \theta =   \frac{1}{cos \theta} } \\  \large \boxed{ {sec}^{2}  \theta =  \frac{1}{ {cos}^{2}  \theta} }

And thus,

\large{  {cos}^{2}  \theta =  \frac{1}{17}}   \\ \large{cos \theta =  \frac{ \sqrt{1} }{ \sqrt{17} } } \\  \large{cos \theta =  \frac{1}{ \sqrt{17} }  \longrightarrow  \frac{ \sqrt{17} }{17} }

Since the given domain is 180° < θ < 360°. Thus, the cosθ < 0.

\large{cos \theta =   \cancel\frac{ \sqrt{17} }{17} \longrightarrow cos \theta =  -  \frac{ \sqrt{17} }{17}}

Then use the Identity of sinθ = tanθcosθ to find the sinθ.

\large{sin \theta = 4 \times ( -  \frac{ \sqrt{17} }{17}) } \\  \large{sin \theta =  -  \frac{4 \sqrt{17} }{17} }

Answer

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4 0
3 years ago
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Answer:

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