Question Video: Using the Sine Rule to Find the Area of a Triangle Inscribed in a Circle Mathematics • Second Year of Secondary School

Video Transcript

𝑀 is the center of a circle, and 𝐴, 𝐵, and 𝐶 are points on the circumference. If 𝐵𝐶 equals 13 centimeters and the measure of angle 𝐶𝑀𝐵 is 84 degrees, find the area of the circle 𝑀, giving the answer to the nearest square centimeter.

We know that the area of a circle is 𝜋𝑟 squared. So really, this problem is about finding the radius of this circle. Let’s begin by putting the information we’ve been given on the diagram. 𝐵𝐶 is 13 centimeters, and the measure of angle 𝐶𝑀𝐵 is 84 degrees. We don’t know the lengths of 𝑀𝐶 or 𝑀𝐵, but they’re each the radius of the circle.

Now, there are numerous different approaches we could take. But one approach is to apply the law of cosines in the triangle 𝐶𝑀𝐵. This states that 𝑎 squared equals 𝑏 squared plus 𝑐 squared minus two 𝑏𝑐 cos 𝐴, where 𝑏 and 𝑐 represent two sides of a triangle and 𝐴 represents the included angle. In our triangle, 𝑎 is 13 centimeters. The angle 𝐴 is 84 degrees. And the two sides which enclose this angle 𝐴 are each the radius of the circle 𝑟.

We can therefore form an equation. 13 squared equals 𝑟 squared plus 𝑟 squared minus two 𝑟 squared cos of 84 degrees. We can solve this equation to find the value of 𝑟 squared, which we’ll then be able to substitute directly into our area formula. Factorizing the right-hand side of our equation by 𝑟 squared, we have 169 equals 𝑟 squared multiplied by two minus two cos of 84 degrees. Dividing through, we have that 𝑟 squared is equal to 169 over two minus two cos 84 degrees. And we’ll keep our value for 𝑟 squared in this exact form.

We can then substitute this value of 𝑟 squared into the area formula and evaluate on a calculator. Rounding our answer, and we have that the area of circle 𝑀 to the nearest square centimeter is 296 square centimeters.

As I mentioned, there are in fact numerous approaches to this problem, which you can try out yourself if you wish. We could’ve applied the law of sines in triangle 𝐶𝑀𝐵. Or we could’ve divided it in half to form two right triangles and then used right-angle trigonometry.