Tangent Of Angle In A Rectangular Parallelepiped
Let's dive into a cool geometry problem involving a rectangular parallelepiped! We're going to figure out how to find the tangent of the angle between two specific lines within this 3D shape. This might sound intimidating, but we'll break it down step-by-step, making it super clear and easy to follow. Geometry can be fun, guys, especially when we tackle it together!
Understanding the Problem: Visualizing the Parallelepiped
Before we jump into calculations, it's crucial to picture what we're dealing with. Think of a rectangular box – that's essentially our rectangular parallelepiped (also called a cuboid). We have vertices labeled ABCDA'B'C'D', where ABCD forms the base, and A'B'C'D' forms the top face. The edges connecting these faces are perpendicular to both bases, which is what makes it a right rectangular parallelepiped.
We're given the lengths of a few key edges: AA' (the height), A'D, and A'B. We need to find the tangent of the angle between the lines A'B' (an edge on the top face) and BD (a diagonal on the bottom face). To find this tangent, we'll need to use our knowledge of 3D geometry, the Pythagorean theorem, and trigonometric relationships. It's like a puzzle, and we have all the pieces – let's put them together!
To really nail this, spend a moment visualizing these lines within the parallelepiped. Imagine A'B' running along the top, and BD cutting across the base. The angle between them is what we're after. Once you have a good mental picture, the calculations will make a lot more sense. Let's get started with the calculations!
Step 1: Finding the Dimensions of the Base
Okay, so we know AA' = 6√3 cm, A'D = 2√43 cm, and A'B = 12 cm. The first step in solving this problem is to determine the dimensions of the rectangular base ABCD. Since ABCDA'B'C'D' is a rectangular parallelepiped, we know that the faces are rectangles. This means that triangles A'B'A, A'D'A are right-angled triangles.
We can use the Pythagorean theorem on triangle A'B'A to find AB. Remember the theorem? It states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. So, in our case:
AB² + AA'² = A'B²
Plugging in the values we know:
AB² + (6√3)² = 12² AB² + 108 = 144 AB² = 36 AB = 6 cm
Now, let's do the same for triangle A'D'A to find AD:
AD² + AA'² = A'D²
Substituting the given values:
AD² + (6√3)² = (2√43)² AD² + 108 = 172 AD² = 64 AD = 8 cm
Great! We've found that AB = 6 cm and AD = 8 cm. These are the lengths of the sides of our rectangular base. Knowing these dimensions is crucial for the next steps in finding the tangent of the angle.
Step 2: Calculating the Length of BD
Now that we have the lengths of AB and AD, we can find the length of the diagonal BD using, you guessed it, the Pythagorean theorem again! This time, we're focusing on the right-angled triangle ABD in the base of the parallelepiped.
BD² = AB² + AD²
We already know AB = 6 cm and AD = 8 cm, so we can plug those values in:
BD² = 6² + 8² BD² = 36 + 64 BD² = 100 BD = 10 cm
Awesome! We've calculated the length of BD to be 10 cm. This diagonal is a key component in understanding the spatial relationship between the lines A'B' and BD, which is what we need to find the angle between them. With BD in our toolkit, we're one step closer to solving the problem. Let's move on to the next step!
Step 3: Visualizing and Constructing the Key Triangle
This is where things get a little more spatial, so stick with me! To find the angle between A'B' and BD, we need to visualize a triangle that involves both of these lines. Since A'B' is parallel to AB (they are opposite sides of a rectangle), we can shift our focus to the angle between AB and BD. However, for a clearer geometric picture, we need to introduce another line.
Let's consider the line parallel to BD that passes through A. We can think of it as sliding BD along the base until it touches point A. Now, imagine a point E on this line such that AE = BD. By connecting the points A, B, and E, we form triangle ABE. This is a crucial triangle because the angle between AB and AE is the same as the angle between A'B' and BD (since A'B' is parallel to AB and AE is parallel to BD). We can denote this angle as ∠BAE, or simply θ.
Furthermore, notice that triangle ABE has sides AB and AE with known lengths (AB = 6 cm and AE = BD = 10 cm). To find the tangent of θ, we'll eventually need to find the length of the third side, BE, or at least the altitude of the triangle corresponding to the base AB or AE. This step is about setting up the right geometric framework, and triangle ABE is our key! Let's calculate the length of BE in the next step.
Step 4: Calculating the Length of BE
To find the length of BE, we'll use the parallelogram law. Think of ABD and our newly constructed segment AE as forming a parallelogram ABDE (since AE is parallel and equal in length to BD). The parallelogram law relates the lengths of the sides of a parallelogram to the lengths of its diagonals.
The parallelogram law states:
2(AB² + AD²) = AE² + BE²
We know AB = 6 cm, AD = 8 cm, and AE = BD = 10 cm. Plugging these values into the equation:
2(6² + 8²) = 10² + BE² 2(36 + 64) = 100 + BE² 2(100) = 100 + BE² 200 = 100 + BE² BE² = 100 BE = 10 cm
So, we've found that BE = 10 cm. This is interesting – triangle ABE has sides AB = 6 cm, AE = 10 cm, and BE = 10 cm. This means triangle ABE is an isosceles triangle (since two sides are equal). Knowing this will help us simplify the final calculation of the tangent of the angle.
Step 5: Determining the Tangent of the Angle
Now comes the grand finale! We've built up all the necessary pieces to find the tangent of the angle between A'B' and BD. Remember, we're looking for the tangent of ∠BAE (which we've called θ) in triangle ABE.
Since triangle ABE is isosceles with AE = BE = 10 cm, we can draw an altitude from E to AB, let's call the point of intersection F. This altitude, EF, bisects AB because it's an isosceles triangle. So, AF = FB = AB / 2 = 6 cm / 2 = 3 cm.
Now, we have a right-angled triangle AFE. We know AE = 10 cm and AF = 3 cm. We can use the Pythagorean theorem again to find the length of EF:
EF² + AF² = AE² EF² + 3² = 10² EF² + 9 = 100 EF² = 91 EF = √91 cm
Finally, we can calculate the tangent of the angle θ (∠EAF) using the definition of tangent in a right-angled triangle:
tan(θ) = Opposite / Adjacent = EF / AF = √91 / 3
Therefore, the tangent of the angle between lines A'B' and BD is √91 / 3. Woohoo! We did it!
Conclusion
Finding the tangent of the angle between lines A'B' and BD in this rectangular parallelepiped was quite the journey! We used a combination of spatial visualization, the Pythagorean theorem, the parallelogram law, and trigonometric relationships. Remember, the key to solving complex geometry problems is to break them down into smaller, manageable steps.
We started by understanding the problem and visualizing the parallelepiped. Then, we systematically calculated the dimensions of the base, the length of BD, and the crucial length BE using the parallelogram law. Finally, we used our knowledge of isosceles triangles and right-angled trigonometry to find the tangent of the angle. Geometry, like any skill, gets easier with practice. So keep exploring, keep visualizing, and keep solving! You got this!