In the right triangle shown, m\angle A = 30\degreem∠A=30°m, angle, A, equals, 30, degree and AB = 12\sqrt{3}AB=12
3

Line AC = 24 units and line BC = 12 units.

Step-by-step explanation: Please refer to the diagram attached for more details.

The right angled triangle ABC has been drawn with angle A measuring 30 degrees and line AB measuring 12√3. To calculate the other two unknown sides AC labelled b, and BC labelled a, we shall use the trigonometric ratios. However, in this scenario, we shall apply the special values of each trigonometric ratio. These are shown in the box on the top right in the picture. The proof is given in the second right angled triangle at the bottom part of the attached picture.

Assume an equilateral triangle with lengths 2 units on all sides and 60 degrees at all angles. Drawing a line perpendicular to the bottom line would divide the top angle into two equal halves (30 degrees each), and the bottom line also would be divided into two equal halves (1 unit each). So the hypotenuse will measure 2 units and the line at the base would measure 1 unit. By using the Pythagoras' theorem, we derive the third side to be √3 (that is x² = 2² - 1², and then x² = 4 - 1,  and then x² = 3 and finally x = √3).

Therefore, in triangle ABC, using angle 30 as the reference angle, to calculate side AC;

Cos 30 = Adjacent/Hypotenuse

Cos 30 = (12√3)/b

b = (12√3)/Cos 30

Where Cos 30 is √3/2

b = (12√3)/√3/2

b = (12√3) * 2/√3

b = 12 * 2

b = 24

To calculate side BC;

Tan 30 = Opposite/Adjacent

Tan 30 = a/(12√3)

Tan 30 * 12√3 = a

Where Tan 30 = 1/√3

(1/√3) * 12√3 = a

12 = a

Therefore, the missing lengths in the right triangle are

AC = 24 units and BC = 12 units