The image shows an acute angled triangle of 30 degrees with sides of 8cm & 5 cm.
A perpendicular has been constructed from point A to the side BC. & the point it meets side BC is marked D.
A circle is drawn through the points A ,D & C.
A tangent is constructed to the circle from point C.
The side it meets AD produced is marked X
Show that $\angle$AXC = $\angle$ACB , Any Ideas on how to begin ?
$\triangle ACX$ is a right triangle with altitude $D.$
$\triangle XDC \sim \triangle CDA ~\sim \triangle XCA$
And that is all you need.