How to calculate angles and areas (circles)- AS Maths

Question

Hi here's the question:

A(-1,-4) and B(6,-5) are points on the circumference of a circle, centre D(3,-1). The tangents at A and B intersect at C. How would I find the angle ACB and the area of ACBD? I have already found the coordinates of C which is (2,-8). Quick replies would be appreciated, thank you!

Edit- this is a Core 1 question so no calculators allowed and also no cosine rule.

Answer 1

Angle $ \theta$ between normals equals angle between tangents.

$$ \tan \theta= \dfrac {m_1-m_2}{1+m_1 m_2} $$

Since you have done everything else this hint should be sufficient.

Area = $ AC \cdot AD $

Answer 2

According the law of cosines \begin{align} \cos(\angle ACB)&=\frac{AC^2+BC^2-AB^2}{2AC\cdot BC}, \end{align} in this case it is 0, hence $ABCD$ is a square since $AD=BD$.