The perimeter of a triangle is 70 cm, its two sides are 24 cm and 32 cm.

The perimeter of a triangle is 70 cm, its two sides are 24 cm and 32 cm. Find the segments into which the bisector of the triangle divides its third side.

AC = P – (AB + BC) = 70 – (24 + 32) = 14 (cm);

cos C = (BC² + AC² – AB²): (2 * BC * AC) = (32² + 14² – 24²): (2 * 32 * 14) = 644: 896 ≈ 0.72;

<C ≈ 46 °;

cos A = (AB² + AC² – BC²): (2 * AB * AC) = (24² + 14² – 32²): (2 * 24 * 14) = -252: 672 ≈ ≈ -0.38;

<A≈ 109 °;

<B ≈ 180 ° – 46 ° – 109 ° ≈ 25 °;

BK is the bisector of angle B; <ABK = <KBC = 1/2 * 25 = 12.5 °; <AKB = 180 ° – 109 ° -12.5 ° ≈58.5 °

AK / sin <ABK = AB / sin <AKB; AK = AB * sin <ABK: sin <AKB;

sin 12.5 ° ≈ 0.21; sin 58.5 ° ≈ 0.45;

AK = 24 * 0.21: 0.84 ≈ 6 (cm); KC ≈ 14 – 6 = 8 (cm).