The two sides of the triangle are 5√3 and 6 cm, and the heights drawn

The two sides of the triangle are 5√3 and 6 cm, and the heights drawn to these sides intersect at an angle of 60 degrees. Find the area of the triangle.

The KOH angle is adjacent to the KOH angle, the sum of which is 180.

Then the angle KOH = (180 – 60) = 120.

Since AH and BK are the heights of the triangle, the angle of РKС and ONС is straight.

In a convex quadrilateral СKOН, the sum of the internal angles is 360, then the angle KCH = (360 – 120 – 90 – 90) = 60.

Then the area of the triangle ABC is equal to: Sавс = АС * ВС * Sin60 / 2 = 5 * √3 * 6 * √3 / 4 = 22.5 cm2.

Answer: The area of the triangle is 22.5 cm2.