A circle has a diameter of 20cm. The line AC is a diameter of the circle. B is a point on the circumference of the circle with AB = 12cm. The point D lies on the arc AC on the opposite side to point B. What is the sine of ∠BDC?

Answer:

27) 0.8

Let the centre of the circle be O.

Here is how the diagram should look like:

AO = CO = 10 cm

Since D is opposite B, BD is also a diameter of the circle, meaning:
BO = DO = 10 cm

We know that triangle ABC is a right-angled triangle. We can find out BC using Pythagoras theorem.

AC² = AB² + BC²
20² = 12² + BC²
400 – 144 = BC²
BC² = 256
BC = 16 cm

We know triangle DBC is also a right-angle triangle with the hypotenuse being AC.

sine∠BDC =    Opposite    = BC = 16
                       Hypotenuse     BD    20

16 ÷ 20 = 0.8