Given a triangle ABC, angle B = 60, side BC = 4cm, and side BA = 5 √ 3, find the area of triangle ABC.

Let’s draw the height of the AK to the side of the BC. Triangle AVK – rectangular with right angle AKB. Angle B is 60 °. If there is an angle of 60 ° in a right-angled triangle, then the leg adjacent to this angle is half the hypotenuse. This means that the VK leg is equal to half of the AB hypotenuse:

BK = AB / 2 = 5√3 / 2 cm.

Find the AK leg according to the Pythagorean theorem (the sum of the squares of the legs is equal to the square of the hypotenuse):

AK² + BK² = AB²;

AK² + (5√3 / 2) ² = 5√3²;

AK² + 18.75 = 75;

AK ² = 75 – 18.75;

AK² = 56.25;

AK = √56.25;

AK = 7.5 cm.

The area of ​​a triangle is equal to the half-product of the base and the height drawn to it:

S = 1/2 * h * ah;

S = 1/2 * AK * BC;

S = 1/2 * 7.5 * 4;

S = 15 cm²

Answer: 15 cm²