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This is not a homework. Just a sudden mathematical spark of my brain prompted me to simply calculate this. enter image description here

In the diagram above the area of the hatched section is 10% of total area if the circle. What the angle CAB should be to satisfy this condition. I need non calculus solution.

So far I have done this... enter image description here

But my final equation cannot be simplified further, or I don't know how to proceed further. What I need is an equation in which the alpha will be the title, so that I can find this alpha for any given area of the hatched section A.

soosai steven
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2 Answers2

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Another approach. From Thales' theorem we know that $\beta=90°$ and $\alpha+\gamma=90°$ Thus $\gamma=90°-\alpha$.

And the triangle $BOC$ is an isosceles triangle. Therefore the angle at $B$ of the triangle $BOC$ is $90°-\alpha$ as well. Then the angle at $0$ of $BOC$ is

$\delta=180°-2\cdot (90°-\alpha)$=$2\alpha$

Now you can use the well known formula for the area of a circular segment:

$A=\frac{r^2}{2}\cdot\left(\frac{\delta \cdot \pi}{180}-\sin (\delta) \right) $

Remark:

You can also use the identity $\boxed{\sin(2\cdot \alpha)=2\cdot \cos(\alpha)\cdot \sin(\alpha)}$.

Your intermediate result $ A=\frac{\alpha \cdot \pi}{180}\cdot r^2-r^2\cdot \cos(\alpha)\cdot \sin(\alpha)$ becomes

$$A=\frac{\alpha \cdot \pi}{180}\cdot r^2-r^2\cdot \frac12\cdot \sin(2\alpha)$$

callculus42
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  • So,how do we write this angle in terms of percentage of the hatched area with respect to the total area of the circle? – soosai steven Jun 12 '16 at 14:26
  • It is not possible to solve directly for $\alpha$. Your equation can only be simplified and then you have to use and an approximation method, for instance the newton raphson method. You was on the right track. See my edit. – callculus42 Jun 12 '16 at 16:08
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Let segment ABC be the given such that [segment ABC] = [yellow] + [orange] = $\dfrac {1}{10}\pi r^2$. At this stage, C is just a point on the minor arc AB.

enter image description here

Construction: Draw the sector OBC with its central angle = $\angle BOC = 36^0 = … = \dfrac {\pi^c}{5}$. Let $x^c$ be the central angle of sector OCA.

Then, [sector OBC] = [pink] + [orange] = $\dfrac {1}{2} r^2 \dfrac {\pi}{5}^c$.

In other words, [pink] = [yellow] and therefore, [pink] + [blue] = [yellow] + [blue].

That is, area of $\triangle OAB$ (edited) = area of sector OAC.

∴ $\dfrac {1}{2} r^2 \sin (x + \dfrac {\pi}{5}) = \dfrac {1}{2} r^2 x^c$.

After simplification, we have $x = \sin (x + \dfrac {\pi}{5})$, which is still in transcendental (but a bit simpler). From Wolframalpha, $x = 0.998435^c$

Mick
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  • Forgive me, I'm not expert in mathematics. I don't understand this, you said C is just a point on minor arc AB, but then in your calculation pi is raised to the power of C. I don't understand it. Second, you had also write triangle QAB, where the Q is in the drawing? – soosai steven Jun 13 '16 at 17:48
  • @soosaisteven That is not an exponent of $c$, that is an alternative notation for radian measure. – Deepak Jun 14 '16 at 03:32
  • @soosaisteven 1) At 1st, C is just any pt on the minor arc AB. Later, because of the given ([segment ABC] = 1/10 [the circle]), I have to create a sector OBC (edited, thanks 4 pointing that out) whose area is also exactly 1/10 [circle]. By this time, C is fixed. 2) In math, besides using degrees in measuring an angle, there is another unit called radian and is short-named as $^c$. The relation between them is $\pi^c = 180^0$. Note that the notation used (in radian) is in accordance of that of the degree. In calculus, the developments of all math processes are based on angles in radians. – Mick Jun 14 '16 at 03:54
  • @soosaisteven Also, $\pi^c = 180^0%$ further means $1^c = 57.xxxxx^0$ approximately. – Mick Jun 14 '16 at 03:57
  • @soosaisteven In calculating [sector OBC], it is (using degree formula) $\pi r^2 \dfrac {36^0}{360^0}$; while, if formula using radians, $\dfrac {1}{2}r^2 ( \dfrac {\pi}{5})^c$. for convenience, sometimes, we just write that without the $^c$. – Mick Jun 14 '16 at 04:10
  • Okay, I understand... Thanks Mick – soosai steven Jun 15 '16 at 10:40