The inscribed circle in triangle ABC touches sides AB, BC and AC at points K L M

The inscribed circle in triangle ABC touches sides AB, BC and AC at points K L M, respectively. Find KL if AM = 2, МС = 3 and angle С = π / 3.

By the property of tangents to the circle drawn from one point, the segments of the tangents are equal:
AM + AK = 2;
CM = CL = 3;
BK = BL = x.

By the cosine theorem:
AB ^ 2 = AC ^ 2 + BC ^ 2 – 2 * AB * BC * cos (π / 3);
(x + 2) ^ 2 = 52 + (x + 3) ^ 2 – 2 * 5 * (x + 3) * (1/2);
x = 5;

From the right-angled triangle MOS:
OM = MC * tan (π / 6) = 3 * (√3 / 3) = √3;

Quadrangle BKLO:
∠ BKO = ∠BLO = 90 °;

BK = BL = 5;
OK = OL = √3;
BO = √5 ^ 2 + (√3) ^ 2 = √28 = 2√7;

The diagonals of the quadrilateral BO and KL are mutually perpendicular.

From the formula for calculating the area of a right-angled triangle:
S = (1/2) a * b and S = (1/2) C * h;

(1/2) KL = BK * OK / BO;

KL = 2 * 5 * √3 / 2√7 = 5√3 / 7;

Answer: 5 √3 / 7