Trigonometry & Inverse Trigonometry
Circle Geometry
Grade 11
Question:
<p>If inside a big circle exactly n (n > 3) small circles, each of radius r, can be drawn in such a way that each small circle touches the big circle and also touches both its adjacent small circles, then the radius of big circle is</p>
<p>(a) \(r\left(1 + \csc\frac{\pi}{n}\right)\)</p>
<p>(b) \(r\frac{1 + \tan\frac{\pi}{n}}{\cos\frac{\pi}{n}}\)</p>
<p>(c) \(r\left(1 + \csc\frac{\pi}{n}\right)\)</p>
<p>(d) \(r\frac{\sin\frac{\pi}{n} + \cos\frac{\pi}{n}}{\sin\frac{\pi}{n}}\)</p>
Step-by-Step Solution
Key Concept: Model the configuration using a regular polygon formed by circle centers and apply the chord-angle relationship.
<p>The centers of n small circles form a regular n-gon on a circle of radius \(R - r\), where R is the radius of the big circle.</p><p>The angle subtended at the center between adjacent small circle centers is \(\frac{2\pi}{n}\).</p><p>The distance between adjacent small circle centers equals \(2r\) (since they touch).</p><p>Using the relation: \(2(R-r)\sin\frac{\pi}{n} = 2r\)</p><p>\(R - r = \frac{r}{\sin\frac{\pi}{n}} = r\csc\frac{\pi}{n}\)</p><p>\(R = r\left(1 + \csc\frac{\pi}{n}\right)\)</p>
Correct Answer: A