4. В прямоугольном треугольнике острый угол равен 60°, а биссектриса этого угла — 8 см. Найдите длину катета, лежащего против этого угла.

Решение:

В прямоугольном треугольнике у нас есть два острых угла. Один из них дан — 60°. Второй угол равен 90° - 60° = 30°.

Биссектриса делит угол пополам. Значит, она делит угол 60° на два угла по 30°.

Рассмотрим треугольник, образованный биссектрисой, катетом и частью гипотенузы. В этом треугольнике:

• Один угол равен 90° (это угол треугольника, из которого проведена биссектриса).
• Второй угол равен 30° (половина угла 60°).
• Третий угол (между биссектрисой и гипотенузой) равен 180° - 90° - 30° = 60°.

Теперь рассмотрим треугольник, образованный биссектрисой, другим катетом и гипотенузой. В этом треугольнике:

• Один угол равен 30° (это второй острый угол исходного треугольника).
• Второй угол равен 30° (половина угла 60°, образованная биссектрисой).
• Следовательно, третий угол (угол между биссектрисой и катетом, лежащим против угла 60°) равен 180° - 30° - 30° = 120°.

В треугольнике, где биссектриса равна 8 см, угол, лежащий напротив катета, равен 30°, а угол, лежащий напротив биссектрисы, равен 60°. Мы можем использовать теорему синусов:

Пусть a — искомый катет, b — второй катет, c — гипотенуза.

Пусть биссектриса l = 8 см делит угол A = 60° на два угла по 30°. Биссектриса проведена из вершины угла A.

В треугольнике, образованном биссектрисой, катетом a и частью гипотенузы, углы равны 90°, 30°, 60°.

Катет a лежит напротив угла 30° (половина угла A).

В этом треугольнике, катет a равен половине гипотенузы, проведенной из вершины угла 90°.

Применим теорему синусов к треугольнику, образованному биссектрисой:

Пусть угол, который делит биссектриса, равен The other acute angle is $$90^\text{o} - 60^\text{o} = 30^\text{o}$$.

The bisector divides the $$60^\text{o}$$ angle into two $$30^\text{o}$$ angles.

Let's consider the triangle formed by the bisector, the leg opposite the 60° angle, and a portion of the hypotenuse. The angles in this triangle are:

• $$90^\text{o}$$ (the right angle of the original triangle)
• $$30^\text{o}$$ (half of the 60° angle)
• $$180^\text{o} - 90^\text{o} - 30^\text{o} = 60^\text{o}$$ (the angle between the bisector and the hypotenuse)

Now consider the triangle formed by the bisector, the other leg (opposite the 30° angle), and the hypotenuse.

Let the leg opposite the 60° angle be $$a$$, the leg opposite the 30° angle be $$b$$, and the hypotenuse be $$c$$. The bisector $$l = 8$$ cm originates from the vertex of the 60° angle.

In the triangle formed by the bisector, the leg $$a$$, and a segment of the hypotenuse, we have angles $$90^\text{o}$$, $$30^\text{o}$$, and $$60^\text{o}$$.

The leg $$a$$ is opposite the $$30^\text{o}$$ angle in this sub-triangle. Therefore, $$a$$ is half of the hypotenuse of this sub-triangle.

Let's use the angle bisector theorem. The ratio of the sides adjacent to the angle is equal to the ratio of the segments the bisector divides the opposite side into. However, this is not directly applicable here as we don't know the hypotenuse lengths of the sub-triangles.

Let's use trigonometry in the smaller triangles formed by the bisector.

Consider the triangle formed by the bisector (length 8), the leg $$a$$, and a part of the hypotenuse. The angles are $$90^\text{o}$$, $$30^\text{o}$$, $$60^\text{o}$$. The side $$a$$ is opposite the $$30^\text{o}$$ angle. So, $$a = \frac{1}{2} h_1$$, where $$h_1$$ is the hypotenuse of this smaller triangle.

Consider the triangle formed by the bisector (length 8), the leg $$b$$, and the other part of the hypotenuse. The angles are $$30^\text{o}$$, $$30^\text{o}$$, $$120^\text{o}$$. This is an isosceles triangle, so the leg $$b$$ is equal to the segment of the hypotenuse. This is incorrect as the leg $$b$$ is not opposite a $$30^\text{o}$$ angle in this sub-triangle relative to the bisector.

Let's restart with a clearer approach.

In a right-angled triangle, let the angles be $$90^\text{o}$$, $$60^\text{o}$$, and $$30^\text{o}$$.

Let the vertex with the $$60^\text{o}$$ angle be A, the vertex with the $$30^\text{o}$$ angle be B, and the vertex with the $$90^\text{o}$$ angle be C.

The bisector of angle A is drawn, and its length is 8 cm. This bisector intersects the opposite side (leg BC) at point D.

The angle bisector divides angle A into two angles of $$30^\text{o}$$ each. So, $$\triangle ABD$$ has angles $$30^\text{o}$$ (at A), $$\theta_1$$, and $$\theta_2$$. $$\triangle ACD$$ has angles $$30^\text{o}$$ (at A), $$\theta_3$$, and $$\theta_4$$.

Let's consider the triangle $$\triangle ACD$$. The angle at A is $$30^\text{o}$$. The angle at C is $$90^\text{o}$$. Therefore, the angle at D (within $$\triangle ACD$$) is $$180^\text{o} - 90^\text{o} - 30^\text{o} = 60^\text{o}$$.

In $$\triangle ACD$$, AD = 8 cm is the hypotenuse. The side AC is opposite the $$60^\text{o}$$ angle, and CD is opposite the $$30^\text{o}$$ angle.

Using trigonometry in $$\triangle ACD$$:

• $$AC = AD \times \tan(60^\text{o}) = 8 \times \frac{\text{opposite}}{\text{adjacent}}$$. No, AD is the hypotenuse.
• $$AC = AD \times \frac{\text{adjacent}}{\text{hypotenuse}} = 8 \times \frac{\text{adjacent}}{\text{hypotenuse}}$$. This is incorrect, AC is a leg.
• $$AC = AD \times \frac{\text{opposite}}{\text{hypotenuse}} \times \text{...?}$$ This is confusing.

Let's use the definition of sine and cosine in the right triangle $$\triangle ACD$$:

• The angle at C is $$90^\text{o}$$.
• The angle at A is $$30^\text{o}$$.
• The angle at D is $$60^\text{o}$$.
• The hypotenuse is AD = 8 cm.
• The leg opposite to angle D (60°) is AC.
• The leg opposite to angle A (30°) is CD.

So, in $$\triangle ACD$$:

• $$AC = AD \times \text{sin}(60^\text{o}) = 8 \times \frac{\text{sqrt(3)}}{2} = 4\text{sqrt(3)}$$.
• $$CD = AD \times \text{sin}(30^\text{o}) = 8 \times \frac{1}{2} = 4$$.

The question asks for the length of the cathetus lying opposite the 60° angle. This is the cathetus AC.

So, the length of the cathetus is $$4\text{sqrt(3)}$$ cm.

Let's verify this. In the original triangle ABC:

• Angle A = 60°, Angle B = 30°, Angle C = 90°.
• AC = $$4\text{sqrt(3)}$$.
• CD = 4.

Now consider $$\triangle ABD$$. Angle at A = $$30^\text{o}$$. Angle at B = $$30^\text{o}$$. Angle at D = $$120^\text{o}$$.

This means $$\triangle ABD$$ is an isosceles triangle with AD = BD. So, BD = 8 cm.

Now we have the lengths of the segments of the leg BC: CD = 4 cm and BD = 8 cm. So, the total length of the leg BC is $$BC = CD + BD = 4 + 8 = 12$$ cm.

Let's check if this is consistent with the original triangle ABC.

In $$\triangle ABC$$:

• $$AC = 4\text{sqrt(3)}$$.
• $$BC = 12$$.
• $$\text{tan}(60^\text{o}) = \frac{BC}{AC} = \frac{12}{4\text{sqrt(3)}} = \frac{3}{\text{sqrt(3)}} = \text{sqrt(3)}$$. This is correct, as $$\text{tan}(60^\text{o}) = \text{sqrt(3)}$$.

Therefore, the cathetus lying opposite the 60° angle is AC, and its length is $$4\text{sqrt(3)}$$ cm.

To express this as a decimal:

$$4\text{sqrt(3)} \text{ cm} \text{ approx } 4 \times 1.732 \text{ cm} = 6.928 \text{ cm}$$.

The problem does not specify the format of the answer, so $$4\text{sqrt(3)}$$ is exact.

The question is: "Найдите длину катета, лежащего против этого угла." (Find the length of the cathetus lying opposite this angle.)