In the figure shown, triangle ABC inscribed in the circle O. If AB = 32, the angle A = 80^{o}, AC = 45. What is the radius of the circle?

Solution:

When a triangle inscribed in a circle, then 2 r = a/sin A = b/sin B = c/sin C.

In which, r is the radius of the circle, a, b and c are three sides of the triangle ABC. In which, a is the side opposite the angle A, b is the side opposite the angle B, and c is the
side opposite the angle C. So, a = BC which is unknown, b = AC = 45, and c = AB = 32.

Apply the Cosine law to the triangle ABC,

a^{2} = b^{2} + c^{2} - 2 b c cos A

= AC^{2} + AB^{2} - 2 × AC × AB × cos80^{o}

= 45^{2} + 32^{2} - 2 × 45 × 32 × 0.1736

= 2549.03

a = 50.5

using formula,

2 r = a / sin A

r = a / (2 sin A)

= 50.5 / (2 × sin 80^{o})

= 50.5 / (2 × 0.9848)

= 25.63

Therefore, the radius of the circle is 25.63. Please watch the video for more details.