The diagram shows a quadrilateral $ABCD$ made up of two triangles. $AB = 20\ \text{cm},\ AD = 12\ \text{cm},\ CD = 32\ \text{cm}.$ Angle $CBD = 90^\circ$ and angle $BCD = 30^\circ$.

1. Identify the correct trigonometric ratio in right-angled triangle $BCD$.

In $\triangle BCD$, $\angle CBD=90^\circ$, so $CD$ is the hypotenuse.

Relative to $\angle BCD=30^\circ$, $BD$ is the opposite side.

$\sin 30^\circ=\frac{BD}{CD}$

2. Substitute the given length and calculate $BD$.

$\sin 30^\circ=\frac{BD}{32}$

$BD=32\sin 30^\circ$

$BD=32\times \frac12=16$

1. Show that triangle $ADB$ is right-angled using the side lengths.

From part (a), $BD=16\text{ cm}$.

$AD^2+BD^2=12^2+16^2=144+256=400$

$AB^2=20^2=400$

Hence,

$AD^2+BD^2=AB^2.$

2. Connect the right angle to the circle with diameter $AB$.

By the converse of Pythagoras' theorem, $\angle ADB=90^\circ$.

Since the angle subtended by a diameter at the circumference is a right angle, point $D$ can lie on the circumference of the circle with diameter $AB$.