Circle question (triangles too)

[Couperin]

Homework Statement

O is the centre of the circle of radius 6cm, and A and B are two points on the circumference such that angle AOB = $$\theta$$ radians.

Show that the length AB is equal to $$\sqrt {72 (1 - cos \theta)}$$

Homework Equations

I think the following might be relevant:

Cosine rule: $$\theta = cos^-1\frac{b^2 + c^2 - a^2}{2bc}$$

Area of sector = $$\frac{1}{2}r^2\theta$$

The Attempt at a Solution

I don't really know where to start here. I think pythagoras is involved somewhere. The radius 6 must therefore be squared, multplied by 2 = 72. But I don't want to work backwards from the given soultion.

Am I missing some theory about chords of circles?

 

[Couperin]

No matter, I solved it.

Treat AOB as a triangle. Apply cosine rule to find AB, rearrange and factorise under the squareroot.

 

[Architeuthis Dux]

I would suggest starting by drawing a diagram of the given information. This will help visualize the problem and make it easier to understand.

From the given information, we know that O is the center of the circle and the radius is 6cm. This means that OA and OB are both equal to 6cm.

Next, we can use the cosine rule to find the length of AB. Using the given formula, we get:

\theta = cos^-1\frac{b^2 + c^2 - a^2}{2bc}

Substituting in the values we know, we get:

\theta = cos^-1\frac{6^2 + 6^2 - AB^2}{2(6)(6)}

Simplifying, we get:

\theta = cos^-1\frac{72 - AB^2}{72}

Now, we can use the area of a sector formula to find the area of the sector OAB. We know that the area of a sector is equal to half the radius squared multiplied by the angle in radians. In this case, the angle in radians is \theta, so we get:

Area of sector OAB = \frac{1}{2}(6)^2\theta = 18\theta

We also know that the area of a sector is equal to the length of the arc multiplied by the radius. In this case, the length of the arc is AB, so we get:

Area of sector OAB = AB(6) = 6AB

Since these two expressions are equal to each other, we can set them equal to each other and solve for AB:

18\theta = 6AB

Dividing both sides by 6, we get:

3\theta = AB

Now, we can substitute this value for AB into our original equation:

\theta = cos^-1\frac{72 - AB^2}{72}

\theta = cos^-1\frac{72 - (3\theta)^2}{72}

\theta = cos^-1\frac{72 - 9\theta^2}{72}

To solve for \theta, we can use the quadratic formula:

\theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

In this case, a = 9, b = 0, and c = -72. Substituting these values, we get: