PART 1

For a triangle ABC, take a point D on side AB such that side CD is orthogonal to side AB. We let \(\angle BAC = \dfrac {\pi}{12}\) and let the lengths of side AB and side AD be \(2 \sqrt{2}\) and \(\sqrt{6}\), respectively. Answer the following questions in the corresponding boxes on the answer sheet. They should be simplified as much as possible.

QUESTION 01

From \(\dfrac {\pi}{12} = \dfrac {\pi}{3} - \dfrac {\pi}{4}\), we have

\(\cos \dfrac {\pi}{12} = \dfrac{\bbox[10px, border: 2px solid red]{[2-1]} + \sqrt{2}}{4}\)

QUESTION 02

The length of side AC is \(\bbox[10px, border: 2px solid red]{[2-2]} - 2 \sqrt{3}\).

QUESTION 03

The square of the length of side BC, \((BC)^2\) , is \(\bbox[10px, border: 2px solid red]{[2-3]} - 32 \sqrt{3}\).

QUESTION 04

Thus, the length of side BC is \(\bbox[10px, border: 2px solid red]{[2-4]} - 2 \sqrt{6}\).