# Thread: Trig Help

1. ## Trig Help

Ok I need some help getting this problem started. I have attached a diagram any help would be appreciated.

A block of metal has a 90 degree notch cut from its lower surface. The notched part rests on a circular cylinder of diamter 2.0 cm. If the lower surface of the part is 1.3 cm above the base plane, how deep is the notch?

I have attached a file that shows the diagram. Any help would be appreciated.

2. Hello, ur5pointos2slo!

A block of metal has a 90° notch cut from its lower surface.
The notched part rests on a circular cylinder of diameter 2.0 cm.
If the lower surface of the block is 1.3 cm above the base plane,
how deep is the notch?
Code:
        *-----------------------*
|                       |
|           A           |
|           *           |
|          /:\          |
|         / : \         |
|        /* * *\        |
|     B *   :   * C     |
|      * 1\ : /1 *      |
|     /*    O    *\     |
*----* *    |    * *----*
1.3      *   |1  *
------------* * *------------
D
$A$ is the peak of the notch.
$\angle BAC = 90^o$ .
(Doesn't look like it in the diagram.)

$O$ is the center of the circle.
Radii $OB = OC = OD = 1$ cm.
$\angle ABO = \angle ACO = 90^o$

$\Delta ABO\text{ and }\Delta ACO$ are isosceles right triangles.
. . Hence: . $AO = \sqrt{2}$

Then: . $AD \:=\:\sqrt{2} + 1$

Therefore, the depth of the notch is: . $(\sqrt{2} + 1) - 1.3 \;=\;\boxed{\sqrt{2}-0.3\text{ cm.}}$

3. Originally Posted by Soroban
Hello, ur5pointos2slo!

$A$ is the peak of the notch.
$\angle BAC = 90^o$ . (Doesn't look like it in the diagram.)

$O$ is the center of the circle.
Radii $OB = OC = OD = 1$ cm.
$\angle ABO = \angle ACO = 90^o$

$\Delta ABO\text{ and }\Delta ACO$ are isosceles right triangles.
. . Hence: . $AO = \sqrt{2}$

Then: . $AD \:=\:\sqrt{2} + 1$

Therefore, the depth of the notch is: . $(\sqrt{2} + 1) - 1.3 \;=\;\boxed{\sqrt{2}-0.3\text{ cm.}}$
Thank you so much that was very helpful.