In triangle ABC it is known that angle C = 90 degrees, CD perpendicular to AB, BC = 3cm, CD = √ 8cm

In triangle ABC it is known that angle C = 90 degrees, CD perpendicular to AB, BC = 3cm, CD = √ 8cm, find the lengths of the sides AB, AC, DB.

The BCD triangle is rectangular. By the Pythagorean theorem, we find the side BD.

BC ^ 2 = BD ^ 2 + CD ^ 2;

BD ^ 2 = BC ^ 2 – CD ^ 2;

BD ^ 2 = 3 ^ 2 – (√8) ^ 2 = 9 – 8 = 1; BD = 1 (cm).

From the triangle BCD, we express the cosine of the angle B. The cosine of the acute angle of a right-angled triangle is the ratio of the leg adjacent to this angle to the hypotenuse.

cos B = BD / BC;

cos B = 1/3.

From the triangle ABC we find the hypotenuse AB from the definition of the cosine of the angle B.

cos B = BC / AB;

AB = BC / cos B;

AB = 3: 1/3 = 9 (cm).

By the Pythagorean theorem, AB ^ 2 = AC ^ 2 + BC ^ 2;

AC ^ 2 = AB ^ 2 – BC ^ 2;

AC ^ 2 = 9 ^ 2 – 3 ^ 2 = 81 – 9 = 72;

AC = √72 = 3√8 (cm).

Answer. AB = 9 cm, AC = 3√8 cm, DB = 1 cm.