In triangle ABC, angle B = 60 degrees, angle C = 30, BC = 2. Find its area.

Consider a triangle ABC. From the problem statement, we know that the angle B = 60 degrees, and the angle C is 30 degrees, respectively.

By the theorem on the sum of the angles of a triangle, we know that the sum of the angles of a triangle in Euclidean space is 180 degrees. Therefore, we can find the angle A:

180 – (60 + 30) = 180 – 90 = 90 degrees.

Therefore triangle ABC is right-angled. So the BC side is the hypotenuse.

The area of ​​a right-angled triangle can be found using the formula:

S = 1/2 * a * b

where a and b are legs of a right-angled triangle. In our case:

a = AC

b = AB

c = BC – hypotenuse

From the property of a right-angled triangle, we know that the leg opposite to an angle of 30 degrees is half the hypotenuse. Consequently:

AB = b = 1/2 * c

By the condition of the problem, c = ВС = 2

Consequently:

b = 1/2 * c = 1/2 * 2 = 1

Now, by the Pythagorean theorem, we can find the second leg:

c ^ 2 = a ^ 2 + b ^ 2

a ^ 2 = c ^ 2 – b ^ 2

a = √ (c ^ 2 – b ^ 2) = √ (2 ^ 2 – 1 ^ 2) = √ (4 – 1) = sqrt 3

Then the area of ​​the triangle is:

S = 1/2 * a * b = 1/2 * √ 3 * 1 = √3 / 2