hello, Bahabeach!
To accomodate cars of most sizes, a parking space needs to contain an 18 ft by 8 ft rectangle.
If a diagonal parking space makes an angle of 68 degrees with the horizontal,
how long are the sides of the parallelogram that contain the rectangle?
Code:
: x :
- *-----------* - - - - - -
\ 22°/ \68°
y\ /8 \
\ / \
* \18
\ \
\ \
\ \
18\ /\
\ / \
\ /8 \
\ / \
*------------*
The "width" of the parallelogram is \(\displaystyle x\).
The "length" of the parallelogram is \(\displaystyle y\,+\,18\).
In the top right triangle, we have: \(\displaystyle \:\cos22^o\,=\,\frac{8}{x}\;\;\Rightarrow\;\;x\,=\,\frac{8}{\cos22^o}\,\approx\,8.6\) ft
We also have: \(\displaystyle \:\tan22^o\,=\,\frac{y}{8}\;\;\Rightarrow\;\;y\,=\,8\cdot\tan22^o\,\approx\,3.2\) ft
Therefore, the sides of the parallelogram are about: \(\displaystyle \,8.6\,\times\,21.2\) ft.
