задача 1Две наклонные AC и AB, проведённые ...

Question

задача 1Две наклонные
AC и

AB, проведённые к плоскости

α, образуют с перпендикуляром

AH углы в
30
°
и
45
°
соответственно. Найди

cos∠ACB, если длина перпендикуляра

AH равна
2√ 6

, а проекции наклонных взаимно перпендикулярны.

Thought

Solving for angle ACB in 3D geometry

I’m working with a 3D geometry problem. Point A is above plane α, and the perpendicular AH from A to the plane intersects at H. Lines AC and AB make angles of 30° and 45° with the perpendicular AH. I need to find the cosine of the angle between lines AC and CB at vertex C, where C and B lie on the plane.

Solving for coordinates and angle calculations

I’m setting up coordinates for points A, B, and C in a 3D space. Point A is above plane α at (0, 0, 2√6). Points B and C lie on the plane z=0. Given the angles between vectors AC and AH, I calculate the cosine of this angle using the formula for dot products. The result gives me that c = h/√3.

Calculating cos ∠ACB

I’ve calculated the lengths of vectors AC and CB, both having magnitude 4√2. Using the dot product of vectors CA = (-2√2, 0, 2√6) and CB = (-2√2, 2√6, 0), the dot product is 8 and the magnitude is 4√2. The cosine of angle ∠ACB = 1/4. Now I’ll explain in Russian.

Crafting solution using right triangles and trigonometry

To solve this, I'll use right triangles AHC and AHB. For each triangle, I know the angle between the inclined line and the perpendicular at A, like ∠CAH = 30°. I’ll find HC and HB using trigonometric relations, like tan and cos, then compute AC and AB. Afterward, I'll consider the projection angle of 90° and use vector-based methods to finalize the angle between projections.

Thought for 34s

Answer

Рассмотрим прямоугольные треугольники $AHC$ и $AHB$ (прямой угол в $H$ , потому что $AH\perp \alpha$ , а $HC,HB\subset\alpha$ ).

Дано:
$\angle CAH=30^\circ,\ \angle BAH=45^\circ,\ AH=2\sqrt6$ , и $HC\perp HB$ .

1) Найдём длины наклонных $AC$ и $AB$

В $\triangle AHC$ :

\cos 30^\circ=\frac{AH}{AC}\ \Rightarrow\ AC=\frac{AH}{\cos 30^\circ} =\frac{2\sqrt6}{\sqrt3/2}=4\sqrt2.

В $\triangle AHB$ :

\cos 45^\circ=\frac{AH}{AB}\ \Rightarrow\ AB=\frac{AH}{\cos 45^\circ} =\frac{2\sqrt6}{\sqrt2/2}=4\sqrt3.

2) Найдём проекции $HC$ и $HB$ , а затем $BC$

\tan 30^\circ=\frac{HC}{AH}\ \Rightarrow\ HC=AH\tan 30^\circ=2\sqrt6\cdot\frac1{\sqrt3}=2\sqrt2.

\tan 45^\circ=\frac{HB}{AH}\ \Rightarrow\ HB=AH\tan 45^\circ=2\sqrt6.

Так как $HC\perp HB$ , то $\triangle BHC$ прямоугольный:

BC=\sqrt{HC^2+HB^2}=\sqrt{(2\sqrt2)^2+(2\sqrt6)^2}=\sqrt{8+24}=\sqrt{32}=4\sqrt2.

3) Косинус $\angle ACB$ по теореме косинусов в $\triangle ACB$

Угол при $C$ противолежит стороне $AB$ , значит:

\cos\angle ACB=\frac{AC^2+BC^2-AB^2}{2\cdot AC\cdot BC} =\frac{32+32-48}{2\cdot (4\sqrt2)(4\sqrt2)} =\frac{16}{64}=\frac14.

\boxed{\cos\angle ACB=\frac14}