\(ABCDEF\) is a regular hexagon. The diagonals \(AC\) and \(AD\) are drawn. (a) Write down angle \(ABC\). [1] (b) Find angle \(BAC\). [2] (c) Find angle \(A...

A regular hexagon has \(6\) equal sides and \(6\) equal interior angles. Its exterior angle is \(360\div 6=60^{\circ}\), so each interior angle is \(180-60=120^{\circ}\).

(a) Angle \(ABC\) is an interior angle of the hexagon:

\(120^{\circ}\) [B1]

(b) Drawing \(AC\) makes triangle \(ABC\), in which \(AB\) and \(BC\) are both sides of the regular hexagon and so are equal. The triangle is therefore isosceles with apex angle \(ABC=120^{\circ}\), and the two equal base angles share what remains:

\((180-120)\div 2\) [M1]

\(=60\div 2=30^{\circ}\) [A1]

(c) Angle \(BCD\) is the hexagon's interior angle at \(C\), which is \(120^{\circ}\), and \(AC\) splits it into angle \(BCA\) and angle \(ACD\). Since angle \(BCA=30^{\circ}\) by the same isosceles reasoning as in part (b):

Angle \(ACD=120-30=90^{\circ}\) [B1]

This right angle is a known feature of the regular hexagon: \(AD\) is a long diagonal through the centre, and it meets the shorter diagonal \(AC\) so that triangle \(ACD\) is right-angled at \(C\).