Step 1

The Sides of AB and AC have unit length

Hence, \(\displaystyle\triangle{A}{B}{C}\) is isosceles with Congruent angles \(\displaystyle\angle{B}\) and \(\displaystyle\angle{C}\) NSNK Since \(\displaystyle{m}{\left(\angle{A}\right)}\) is given as 60 this means that,

\(\displaystyle{m}{\left(\angle{B}\right)}+{m}{\left(\angle{C}\right)}={120}\)

and so, \(\displaystyle{m}{\left(\angle{B}\right)}+{m}{\left(\angle{C}\right)}={60}\)

Therefore \(\displaystyle\triangle{A}{B}{C}\) is equilateral

Equilateral triangles have three lines of reflective symmetry in which the lines joining each vertex to the midpoint of the opposite side.

This means that CD is a line of symmetry for \(\displaystyle\triangle{A}{B}{C}\) and so CD is perpendicular to AB.

Step 2

By applying Pythagorean Theorem to the right triangle \(\displaystyle\triangle{A}{D}{C}\)

\(\displaystyle{\left|{A}{D}\right|}{2}+{\left|{C}{D}\right|}{2}={\left|{A}{C}\right|}{2}\)

We Know that D is the midpoint of AB.

B is on the unit circle,

Hence \(\displaystyle{\left|{A}{B}\right|}={1}\) and \(\displaystyle{\left|{A}{D}\right|}={12}\)

Since C is unit circle, We have \(\displaystyle{\left|{A}{C}\right|}={1}\)

Plugging the values for \(\displaystyle{\left|{A}{D}\right|}\) and \(\displaystyle{\left|{A}{C}\right|}\) into the formula gives,

\(\displaystyle{\left|{C}{D}\right|}={3}\sqrt{{{2}}}\). Since CD is perpendicular to AB .

Therefore, \(\displaystyle{C}{\left({12},\ {3}\sqrt{{{2}}}\right)}\)

\(\displaystyle{{\sin{{60}}}^{{\circ}}=}{3}\sqrt{{{2}}}\) and \(\displaystyle{{\cos{{60}}}^{{\circ}}=}{12}\)

The Sides of AB and AC have unit length

Hence, \(\displaystyle\triangle{A}{B}{C}\) is isosceles with Congruent angles \(\displaystyle\angle{B}\) and \(\displaystyle\angle{C}\) NSNK Since \(\displaystyle{m}{\left(\angle{A}\right)}\) is given as 60 this means that,

\(\displaystyle{m}{\left(\angle{B}\right)}+{m}{\left(\angle{C}\right)}={120}\)

and so, \(\displaystyle{m}{\left(\angle{B}\right)}+{m}{\left(\angle{C}\right)}={60}\)

Therefore \(\displaystyle\triangle{A}{B}{C}\) is equilateral

Equilateral triangles have three lines of reflective symmetry in which the lines joining each vertex to the midpoint of the opposite side.

This means that CD is a line of symmetry for \(\displaystyle\triangle{A}{B}{C}\) and so CD is perpendicular to AB.

Step 2

By applying Pythagorean Theorem to the right triangle \(\displaystyle\triangle{A}{D}{C}\)

\(\displaystyle{\left|{A}{D}\right|}{2}+{\left|{C}{D}\right|}{2}={\left|{A}{C}\right|}{2}\)

We Know that D is the midpoint of AB.

B is on the unit circle,

Hence \(\displaystyle{\left|{A}{B}\right|}={1}\) and \(\displaystyle{\left|{A}{D}\right|}={12}\)

Since C is unit circle, We have \(\displaystyle{\left|{A}{C}\right|}={1}\)

Plugging the values for \(\displaystyle{\left|{A}{D}\right|}\) and \(\displaystyle{\left|{A}{C}\right|}\) into the formula gives,

\(\displaystyle{\left|{C}{D}\right|}={3}\sqrt{{{2}}}\). Since CD is perpendicular to AB .

Therefore, \(\displaystyle{C}{\left({12},\ {3}\sqrt{{{2}}}\right)}\)

\(\displaystyle{{\sin{{60}}}^{{\circ}}=}{3}\sqrt{{{2}}}\) and \(\displaystyle{{\cos{{60}}}^{{\circ}}=}{12}\)