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A and B are two points on the circumference of a circle with centre O. C is a point on OB such that AC $\perp OB$. AC = 12 cm. BC = 5 cm. Calculate the size of $\angle AOB$, marked $\theta$ on the diagram.

Angle in a triangle within a circle

The answer given in the textbook is $45.2 ^\circ$ (1dp)

Note: This is not a homework question. I'm just doing maths for my own interest.

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3 Answers 3

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let radius = $r$, then in triangle ACO, using Pythagoras:

$$\begin{align} AO^2 &= AC^2+CO^2\\ \\ r^2 &= 12^2+(r-5)^2\\ \\ 144+25-10r &= 0\\ \\ r &= 16.9\\ \end{align}$$

In triangle ACO, $$\begin{align} \sin\theta &= \dfrac{12}{r}\\ \\ \sin\theta &= \dfrac{12}{16.9}\\ \\ \theta &= 45.2^{\circ} (1dp)\\ \end{align}$$

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Radius $AO=BO=x$ CM(say)

So, $OC=OB-BC=x-5$ i.e., $x\ge5$ and $AC=12$

We have $AO^2=OC^2+AC^2$

$\angle AOB=\arctan\dfrac{12}{x-5}$

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  • $\begingroup$ You didn't calculate $x$. $\endgroup$ Commented Dec 19, 2014 at 8:27
  • $\begingroup$ @ZubinMukerjee, I left it for Blakes7 $\endgroup$ Commented Dec 19, 2014 at 8:30
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Draw line AB, and observe that ABC is a (5,12,13) right-angled triangle. Since AB is a chord, the bisector of angle AOB also bisects AB, call this point T. Now triangle OTB is similar to triangle ACB. Therefore angle AOB is:

2 * arctan(5/12)

Which a handy calculator says is 45.2 degrees.

(I'm surprised someone marked this down: I did get the answer, which one other answer didn't, and I think it's neater than the other answer, since you need almost no calculation.)

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  • $\begingroup$ I didn't down vote but I am having trouble intuitively seeing why OTB is similar to ACB... $\endgroup$
    – Chris
    Commented Dec 19, 2014 at 13:39
  • $\begingroup$ The line AB and its midpoint T aren't in the diagram, and I don't have the means to draw them, but then: angles TBO and ABC are coincident, and angles OTB and BCA are right angles. So angles are all the same. $\endgroup$ Commented Dec 19, 2014 at 13:46
  • $\begingroup$ I think this user down votes answers that don't meet his exact criteria or expected format. $\endgroup$ Commented Jan 4, 2015 at 8:46

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