Question

Suppose another box has a base with dimensions 8 by 6 and diagonal that forms a${70}^{\circ }$angle with a diagonal of a base. Show that the ration of the volumes of the two boxes is $\frac{\tan {35}^{\circ }}{\tan {70}^{\circ }}$.

Short answer

The volume of two boxes is$\frac{\tan {35}^{\circ }}{\tan {70}^{\circ }}$ .

Step-by-step solution

Step 1. Given information.

A diagonal of a box forms a${35}^{\circ }$ angle with a diagonal of the base.

Another diagonal of a box forms a${70}^{\circ }$ angle with a diagonal of the base.

Step 2. Determine the height.

Use Pythagoras theorem,

$\begin{array}{c}\sqrt{{\left(8\right)}^2+{\left(6\right)}^2}=d\\ d=\sqrt{64+36}\text{[square]}\\ d=10\text{[simplify]}\end{array}$

Now, the height of the boxes is,

$\begin{array}{c}\tan {35}^{\circ }=\frac{h}{10}\\ h=10\tan {35}^{\circ }\text{[crossmultiply]}\end{array}$

And

$\begin{array}{c}\tan {70}^{\circ }=\frac{h}{10}\\ h=10\tan {70}^{\circ }\text{[crossmultiply]}\end{array}$

Step 3. Determine the ratio of volume.

The volume of the box is:

$\begin{array}{c}\frac{v_1}{v_2}=\frac{8\times 6\times 10\tan {35}^{\circ }}{8\times 6\times 10\tan {70}^{\circ }}\\ \frac{v_1}{v_2}=\frac{\tan {35}^{\circ }}{\tan {70}^{\circ }}\end{array}$

This is the required expression.