In triangle ABC, angle B = 90 *, angle C = 60 *, BC = 2 cm. Point D is laid on the side AC so that angle ABD = 30

In triangle ABC, angle B = 90 *, angle C = 60 *, BC = 2 cm. Point D is laid on the side AC so that angle ABD = 30 * a) Find the length of the segment AD. b) Prove that the perimeter of triangle ABC is less than 10 cm.

Given:
right-angled triangle ABC,
angle B = 90 *,
angle C = 60 *,
BC = 2 cm,
point D belongs to AC,
angle ABD = 30 *.
a) Find the length of the segment AD.
b) Prove that the perimeter of triangle ABC is less than 10 cm.
a) Solution:
1) Consider a triangle ВСD. Angle ВСD = angle В – angle ABD;
Angle ВСD = 90 – 30 = 60 *;
angle CDA = 180 – 60 – 60 = 60 *.
Therefore, the BCD triangle is equilateral.
ВС = СD = ВD = 2 cm
2) Consider a right-angled triangle ABC. Angle A = 180 – 90 – 60 = 30 *. Then triangle ВDА is isosceles.
BD = DA = 2 cm;
3) CA = CD + DA;
CA = 2 + 2 = 4 cm;
b) By the Pythagorean theorem
ВD ^ 2 = CA ^ 2 – BC ^ 2;
BD ^ 2 = 16 – 4;
BD ^ 2 = 12;
BD = 2√3.
P ABC = 2 + 4 + 2√3 = 8√3 cm <10 cm. Proved.
Answer: 4 centimeters.