Trigonometry three dimensional question

**lalji** · April 26th 2008, 04:58 AM

I need help to solve:
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 at the back of the text book is 16 degrees 15 minutes.

**Mathstud28** · April 26th 2008, 08:01 AM

Originally Posted by lalji

I need help to solve:
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 at the back of the text book is 16 degrees 15 minutes.

Maybe it is the three dimensions that are screwing you up...I say think it more two dimensionally to start...and here is a big hint...PERPENDICULAR height...and perp. is synonomous with what?

**Moo** · April 26th 2008, 08:28 AM

Hello,

Take a look at this picture. OPQR represents the 12cm high box. ABCD is a view from the side of the cylinder.

According to the text,
$RQ=OP=12 cm$

$AB=CD=2 \cdot 4=8 cm$

$AC=BD=15 cm$

And you're looking for angle $DAR$ .

Is it more helpful ?

**Soroban** · April 26th 2008, 09:59 AM

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 $ABCD\!: CB = 8,\;AB = 15,\;PQ = 12$

The box is $PQRS\!:\;PQ = 12$

Let $x = BQ$
Let $\theta = \angle BCQ$
. . Note that $\angle ABP = \theta$

In right triangle $BCQ\!:\;QC = \sqrt{64-x^2}$
. . $\cos\theta \,=\,\frac{\sqrt{64-x^2}}{8}\;\;{\color{blue}[1]}$

In right triangle $APB\!:\;PB \,=\,12-x$
. . $\cos\theta \:=\:\frac{12-x}{15}\;\;{\color{blue}[2]}$

Equate [1] and [2]: . $\frac{\sqrt{64-x^2}}{8} \;=\;\frac{12-x}{15}\quad\Rightarrow\quad 15\sqrt{64-x^2} \;=\;8(12-x)$

Square both sides: . $225(64-x^2) \;=\;64(144-24x + x^2)$

. . which simplifies to: . $289x^2 - 1536x - 5184 \;=\;0$

. . and has the positive root: . $x \;=\;\frac{1536 + \sqrt{6,352,000}}{578} \;=\;7.657439446$

Then: . $\sin\theta \;=\;\frac{x}{8} \;=\;\frac{7.657439446}{8} \;=\;0.957179931$

. . $\theta \;=\;\sin^{-1}(0.957179931) \;=\;73.17235533^o \;\approx\;\boxed{73^o10'}$

**Moo** · April 26th 2008, 10:04 AM

I've got some difficulties, how can we know, by reading the text, what angle we are looking for ? And how can this physically be possible if D is not on RS ??

**Gusbob** · April 26th 2008, 04:44 PM

When I saw this question, I knew I saw it before. It was in my textbook last year. Here is the picture the book provided, and the answer at the back of the book is 16 degrees 50 minutes, not 16 degrees 15 minutes. I think Soroban had a correct answer, but the book is asking for the angle of the other side. So:

$180^o$ (straight angle) - $90^o$ (angle between base and length of cylinder) - $73^o10'$ (angle 'on the other side' Soroban found) = $\theta$

$\theta$ = $16^o 50'$

**lalji** · April 26th 2008, 06:30 PM

Thanks everyone especially Soroban, you are genius. Two dimensional diagram is very helpful to understand. Thanks Gusbob for posting the diagram from the text book, I did not have scanner to scan the diagram.

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