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So this is my homework:

enter image description here

$PQ=2$ and $PQ$ is also the radius of the half-circle, $QR$ is parallel to $PS$, $P$ is the origin,

How do we express the circumference in terms of $\theta$? I still can't imagine how the size $\theta$ affects the circumference. The question asks to prove that $K = 2 + 6 \cos(\theta) + 2 \sin (\theta)$ where $K$ is the circumference of this half circle.

Thank you very much!

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    $\begingroup$ What is $K$? I don't see it specified anywhere? $\endgroup$ Oct 17, 2020 at 2:35
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    $\begingroup$ I don’t see a definition for $K$ in the image. $\endgroup$ Oct 17, 2020 at 2:35
  • $\begingroup$ You will get a better response from mathSE reviewers if you edit your query to show your work. What have you tried? Where are you specifically having trouble? $\endgroup$ Oct 17, 2020 at 2:36
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    $\begingroup$ To link $\theta$ with the circumference, I have to ask you some questions about your math background. What is the circumference of a circle of radius 2? Do you measure angles in degrees or radians? Do you understand the coordination between arc length (which is actually a fraction of 1 complete revolution = 1 complete circumference) and the angle which may be construed to be a fraction of 360 degrees or a fraction of $2\pi$ radians? It is difficult to help without knowing your background here. $\endgroup$ Oct 17, 2020 at 2:39
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    $\begingroup$ Just checking: is $P$ not the origin, and should the question say $QR$ is parallel to $PS$? Make sure you check your question so that it makes sense to someone reading it for the first time. The more context you include, the better your question. $\endgroup$
    – Toby Mak
    Oct 17, 2020 at 3:01

1 Answer 1

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enter image description here

The circumference, or the perimeter of $PQRS$ in terms of the given radius $|PQ|=r=2$ and the value of angle $\theta$ can be found as follows:

\begin{align} K&=|PQ|+|QR|+|RS|+|PS| \tag{1}\label{1} ,\\ |QR|&=2\,|QT| \tag{2}\label{2} ,\\ |QT|&=|PS|=r\cos\theta \tag{3}\label{3} ,\\ |RS|&=r\,\sin\theta \tag{4}\label{4} , \end{align}

so, indeed, we have \begin{align} K&=r+2r\cos\theta+r\,\sin\theta+r\cos\theta =r+3r\cos\theta+r\,\sin\theta \\ &=2+6\cos\theta+2\,\sin\theta \tag{5}\label{5} . \end{align}

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  • $\begingroup$ thank you so much, sir! $\endgroup$
    – Hafizh
    Oct 17, 2020 at 12:15

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