Hello, lalji!
I don't agree with their answer . . .
A cylinder with radius 4 cm and perpendicular height 15 cm
is tilted so that it will just fit inside a 12 cm high box.
At what angle must it be tilted?
Answer given: 16 degrees 15 minutes. Code:
P A S
_ * - - - * - - - - *
: | * * |
: | * *D |
: | * * |
12-x| * 15 * |
: | * * |
: | * * |
: |* * |
-B* 8 * |
x | * * |
- * - - * - - - - - *
Q C R
The cylinder is $\displaystyle ABCD\!: CB = 8,\;AB = 15,\;PQ = 12$
The box is $\displaystyle PQRS\!:\;PQ = 12$
Let $\displaystyle x = BQ$
Let $\displaystyle \theta = \angle BCQ$
. . Note that $\displaystyle \angle ABP = \theta$
In right triangle $\displaystyle BCQ\!:\;QC = \sqrt{64-x^2}$
. . $\displaystyle \cos\theta \,=\,\frac{\sqrt{64-x^2}}{8}\;\;{\color{blue}[1]}$
In right triangle $\displaystyle APB\!:\;PB \,=\,12-x$
. . $\displaystyle \cos\theta \:=\:\frac{12-x}{15}\;\;{\color{blue}[2]}$
Equate [1] and [2]: .$\displaystyle \frac{\sqrt{64-x^2}}{8} \;=\;\frac{12-x}{15}\quad\Rightarrow\quad 15\sqrt{64-x^2} \;=\;8(12-x)$
Square both sides: .$\displaystyle 225(64-x^2) \;=\;64(144-24x + x^2)$
. . which simplifies to: .$\displaystyle 289x^2 - 1536x - 5184 \;=\;0$
. . and has the positive root: .$\displaystyle x \;=\;\frac{1536 + \sqrt{6,352,000}}{578} \;=\;7.657439446$
Then: .$\displaystyle \sin\theta \;=\;\frac{x}{8} \;=\;\frac{7.657439446}{8} \;=\;0.957179931$
. . $\displaystyle \theta \;=\;\sin^{-1}(0.957179931) \;=\;73.17235533^o \;\approx\;\boxed{73^o10'} $