Question

find the angles that the diagonal of a rectangular parallelepiped 2 in. by 3 in. by 4 in. makes with the faces.

Answer 1

Answer on Question#39942 – Math - Geometry

Question.

Find the angles that the diagonal of a rectangular parallelepiped 2 in. by 3 in. by 4 in. makes with the faces.

Solution:

a) We have

$$\tan \alpha = \frac {2}{x},$$ $$\tan \alpha = \frac {2}{\sqrt {3 ^ {2} + 4 ^ {2}}} = \frac {2}{5},$$ $$\alpha = \tan^ {- 1} \left(\frac {2}{5}\right) \approx 0.38051 \text{ (radians)}$$

b) We have

$$\tan \beta = \frac {3}{y},$$ $$\tan \beta = \frac {3}{\sqrt {2 ^ {2} + 4 ^ {2}}} = \frac {3}{\sqrt {2 0}} = \frac {3}{2 \sqrt {5}},$$ $$\beta = \tan^ {- 1} \left(\frac {3}{2 \sqrt {5}}\right) \approx 0.59087 \text{ (radians)}$$

c) We have

$$\tan \gamma = \frac {4}{y},$$ $$\tan \gamma = \frac {4}{\sqrt {2 ^ {2} + 3 ^ {2}}} = \frac {4}{\sqrt {1 3}}.$$ $$\gamma = \tan^ {- 1} \left(\frac {4}{\sqrt {1 3}}\right) \approx 0. 8 3 7 2 2 (r a d i a n s)$$

Answer:

$$\alpha \approx 0. 3 8 0 5 1 (r a d i a n s)$$ $$\beta \approx 0. 5 9 0 8 7 (r a d i a n s)$$ $$\gamma \approx 0. 8 3 7 2 2 (r a d i a n s)$$