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Math Help - Trig Help

  1. #1
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    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.
    Attached Thumbnails Attached Thumbnails Trig Help-notch.jpg  
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  2. #2
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    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.}}

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  3. #3
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    Quote Originally Posted by Soroban View Post
    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.
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