Using the technique in the model above, find the missing segments in this 30°-60°-90° right triangle.BC = 2CD =

Consider right triangle BCD: angle D is right and angle B has measure 60°, then angle C has measure 180°-90°-60°=30°. BC is the hypotenuse of this triangle. In special 30°-60°-90° right triangle the leg that lies opposite to the angle 30° is half of hypotenuse. So, BD=BC/2=2/2=1.Consider right triangle ABC: angle C is right, angle B has measure 60° and angle A has measure 30°. AB is the hypotenuse of this triangle. In special 30°-60°-90° right triangle the leg that lies opposite to the angle 30° is half of hypotenuse. So, BC=AB/2, therefore AB=2BC=2·2=4 .The hypotenuse AB consists of two parts: AD and BD. Since AB=4, BD=1, you have that AD=AB-BD=4-1=3.The height of right triangle drawn to the hypotenuse is the geometrical mean of the previous parts:[tex] CD^2=AD\cdot BD,\\ CD^2=3\cdot 1=3,\\ CD=\sqrt{3} [/tex].Consider right triangle ACD: angle D is right and angle A has measure 30°, then angle C has measure 180°-90°-30°=60°. AC is the hypotenuse of this triangle. In special 30°-60°-90° right triangle the leg that lies opposite to the angle 30° is half of hypotenuse. So, CD=AC/2, AC=2·CD=2·√3=2√3.Answer: AB=4 (hypotenuse), BD=1 and AD=3 (legs projections on the hypotenuse), CD=√3 (height to the hypotenuse), AC=2√3 and BC=2 (legs).