In triangle ABC, angle A = 60 degrees AB = 4, AC = 5. Find the length of the bisector AL of the triangle.

СL and L В – segments into which the bisector divides the side of the triangle ABC, according to the property of the bisector, AC \ AB = CL \ L B ⇒ 5 \ 4 = CL \ L B. ⇒ CL = 5x, LB = 4x, and using the cosine theorem it turns out, CB² = AC² + AB² – 2 * AB * AC * cosine of angle 60. (CB = 5x + 4x = 9x) ⇒ 81x² = 25 + 16 – 2 * 5 * 4 * 1 \ 2 ⇒
81x² = 21 ⇒ x² = ⇒ x = √21 \ 9
С L = 5 * √21 \ 9 = 5√21 \ 9, LВ = 4 * √21 \ 9 = 4√21 \ 9
АL = √ (АС * АВ – СL * LВ) ⇒ АL = √ (5 * 4 – 5√21 \ 9 * 4√21 \ 9) = √ (1200 \ 81) = 20√3 \ 9
A L = 20√3 \ 9