In triangle ABC, angle C is 90, Ab = 5, tgA = √ (1/10). Find BC

Given:

Right-angled triangle ABC;

Angle C = 90 degrees;

AB = 5;

tg A = √ (1/10);

Let’s find the aircraft.

Solution:

In triangle ABC, angle C is 90 degrees.

AB – hypotenuse;

AC, BC – hypotenuse.

According to the Pythagorean formula:

AB ^ 2 = AC ^ 2 + BC ^ 2;

5 ^ 2 = AC ^ 2 + BC ^ 2;

Since, tg A = AC / BC = 1 / √10, then we get: AC = tg A * BC = 1 / √10 * BC.

Then:

5 ^ 2 = (1 / √10 * BC) ^ 2 + BC ^ 2;

25 = 1/10 * BC ^ 2 + BC ^ 2;

25 * 10 = 1/10 * 10 * BC ^ 2 + 10 * BC ^ 2;

250 = BC ^ 2 + 10 * BC ^ 2;

ВС ^ 2 * (1 + 10) = 250;

BC ^ 2 * 11 = 250;

BC ^ 2 = 250/11;

BC = √ (250/11);

BC = 5 * √ (10/11).