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

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.

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