Question

Помогите хотя бы рисунок нарисовать, пожалуйста В правильной треугольной пирамиде SABC, боковое ребро которой равно стороне основания, точка K - середина ребра SB, точка M - середина ребра BC. Найдите косинус угла между прямыми AK и SM.

Answer 1

Ответ:

$\dfrac{1}{6}$

Объяснение:

Прямые АК и SM - скрещивающиеся.

Пусть Н - середина МВ, тогда КН - средняя линия ΔSMB, КН║SM.

∠(АК; SM) = ∠(AK: KH) = ∠AKH = α - искомый угол.

Пусть а - ребро пирамиды (все ребра равны).

$AK=SM=\dfrac{a\sqrt{3}}{2}$  как высоты (медианы) равностороннего треугольника.

$KH=\dfrac{1}{2}SM=\dfrac{a\sqrt{3}}{4}$   (средняя линия треугольника SMB)

АН выразим из ΔАВН по теореме косинусов:

$AH^2=BA^2+BH^2-2\cdot BA\cdot BH\cdot \cos\angle B$

∠B = 60°, так как ΔАВС правильный.

$AH^2=a^2+\dfrac{a^2}{16}-2\cdot a\cdot \dfrac{a}{4}\cdot \dfrac{1}{2}=\dfrac{17a^2}{16}-\dfrac{a^2}{4}=\dfrac{13a^2}{16}$

$AH=\dfrac{a\sqrt{13}}{4}$

Из треугольника АКН по теореме косинусов:

$AH^2=AK^2+KH^2-2\cdot AK\cdot KH\cdot \cos\alpha$

$\cos\alpha =\dfrac{AK^2+KH^2-AH^2}{2\cdot AK\cdot KH}$

$\cos\alpha =\dfrac{\dfrac{3a^2}{4}+\dfrac{3a^2}{16}-\dfrac{13a^2}{16}}{2\cdot \dfrac{a\sqrt{3}}{2}\cdot \dfrac{a\sqrt{3}}{4}}$

$\cos\alpha =\dfrac{2a^2}{16}:\dfrac{3a^2}{4}=\dfrac{2a^2}{16}\cdot \dfrac{4}{3a^2}$

$\boldsymbol{\cos\alpha =\dfrac{1}{6}}$

Answer 2

Finding the Cosine of the Angle between AK and SM in a Right Triangular Pyramid

To find the cosine of the angle between the lines AK and SM in the right triangular pyramid SABC, we can use the properties of similar triangles and the definition of the cosine function.

Let's break down the problem step by step:

1. Identify the given information: - SABC is a right triangular pyramid. - The lateral edge of the pyramid (SB) is equal to the length of the base (BC). - Point K is the midpoint of edge SB. - Point M is the midpoint of edge BC.

2. Visualize the pyramid: - Draw a right triangular pyramid with vertices S, A, B, and C. - Label the sides and edges as mentioned in the problem statement.

3. Identify the relevant triangles: - Triangle AKS: This triangle includes the line AK and shares an angle with the desired angle. - Triangle SMC: This triangle includes the line SM and shares an angle with the desired angle.

4. Use the properties of similar triangles: - Since AK is a median of triangle SBC, it divides the base BC into two equal parts. - Similarly, SM is a median of triangle ABC, dividing the base BC into two equal parts. - Therefore, triangles AKS and SMC are similar.

5. Find the cosine of the angle between AK and SM: - The cosine of an angle can be found using the formula: cos(angle) = adjacent side / hypotenuse. - In triangle AKS, the adjacent side is AK, and the hypotenuse is AS. - In triangle SMC, the adjacent side is SM, and the hypotenuse is SC. - Since triangles AKS and SMC are similar, the ratios of the corresponding sides are equal. - Therefore, cos(angle) = AK / AS = SM / SC.

6. Substitute the given information: - We know that SB = BC, and since K is the midpoint of SB, SK = KB = 0.5 * SB. - Similarly, MC = 0.5 * BC. - Since the lateral edge SB is equal to the base BC, we can substitute SB for BC in the above equations. - Therefore, SK = KB = 0.5 * SB = 0.5 * BC = 0.5 * SC. - Similarly, MC = 0.5 * BC = 0.5 * SB = 0.5 * SC.

7. Calculate the cosine of the angle: - Since AK = SK and SM = MC, we can rewrite the equation as cos(angle) = SK / AS = MC / SC. - Since SK = MC = 0.5 * SC, we can substitute these values into the equation. - Therefore, cos(angle) = 0.5 * SC / AS.

8. Simplify the equation: - To simplify the equation further, we need additional information about the lengths of the sides AS and SC. - Without this information, we cannot determine the exact value of the cosine of the angle.

In conclusion, to find the cosine of the angle between AK and SM in the right triangular pyramid SABC, we need additional information about the lengths of the sides AS and SC.