In a triangle abc, side bc forms an angle with the base ac equal to 30 degrees, and the height drawn from the vertex b divides

In a triangle abc, side bc forms an angle with the base ac equal to 30 degrees, and the height drawn from the vertex b divides the bases into segments, ad = 12cm, dc = 5√3, find the sides

Consider a triangle BCD. BDC angle = 90 °, BCD angle = 30 °. cos (BCD) = cos (30 °) = 1/2 * √3 CD / BC = 5 * √3 / BC => BC = 10. By the theorem of cosines from triangle ABC: AB ^ 2 = AC ^ 2 + BC ^ 2-2 * AC * BC * cos (ACB) AB ^ 2 = (12 + 5 * √3) ^ 2 + 100- (12+ 5 √3) * 10 √ 3. AB ^ 2 = 169 => AB = 13.
Answer: AB = 13, BC = 10.