(a) Substitute the equation of the tangent \(y = mx + 10\) into the circle's equation \(x^2 + y^2 = 20\):

\(x^2 + (mx + 10)^2 = 20\)

\(x^2 + (mx)^2 + 20mx + 100 = 20\)

\((1 + m^2)x^2 + 20mx + 80 = 0\)

Use the discriminant \(b^2 - 4ac = 0\) for tangency:

\((20m)^2 - 4(1 + m^2)80 = 0\)

\(400m^2 - 320 - 80m^2 = 0\)

\(320m^2 = 320\)

\(m^2 = 4\)

\(m = \pm 2\)

(b) For \(m = 2\), the tangent is \(y = 2x + 10\). Substitute into the circle's equation:

\(x^2 + (2x + 10)^2 = 20\)

\(x^2 + 4x^2 + 40x + 100 = 20\)

\(5x^2 + 40x + 80 = 0\)

\(x = -4\) or \(x = 4\)

For \(x = -4\), \(y = 2(-4) + 10 = 2\)

For \(x = 4\), \(y = 2(4) + 10 = 2\)

Coordinates are \((-4, 2)\) and \((4, 2)\).

(c) Use the cosine rule in triangle \(BCD\):

\(BC = 8, BD = \sqrt{(\sqrt{20} + 4)^2 + 2^2}, CD = \sqrt{(\sqrt{20} - 4)^2 + 2^2}\)

\(64 = 80 - 16\sqrt{5} \cos BDC\)

\(\cos BDC = \frac{5}{5}\)

\(\angle BDC = 63.4^\circ\)