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Math Help - Similar Triangles help

  1. #1
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    Similar Triangles help

    To find the height of a tree, Sarah placed a mirror on the ground 15 m from the base of a tree. She walked backward until she could see the top of the tree in the centre of the mirror. At that position she was 1.2m from the mirror and her eyes were 1.4m from the ground.
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  2. #2
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    Quote Originally Posted by sinjid9 View Post
    To find the height of a tree, Sarah placed a mirror on the ground 15 m from the base of a tree. She walked backward until she could see the top of the tree in the centre of the mirror. At that position she was 1.2m from the mirror and her eyes were 1.4m from the ground.
    Hi sinjid9,

    The triangle from the mirror to the tree to the base of the tree
    is a magnified version of the triangle from the mirror to Sarah's eyes to her feet,
    if the tree is taller than her, which the dimensions confirm.

    You may write

    tan\theta=\frac{h}{15}=\frac{1.4}{1.2}

    multiply both sides by 15 to find h

    Or

    \frac{height1}{base1}=\frac{height2}{base2}\ \Rightarrow\ \frac{1.4}{1.2}=\frac{tree\ height}{15}
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  3. #3
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    can you send me a diagram of this question?
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  4. #4
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    Hello, sinjid9!

    To find the height of a tree, Sarah placed a mirror on the ground 15 m from the base of a tree.
    She walked backward until she could see the top of the tree in the centre of the mirror.
    At that position she was 1.2m from the mirror and her eyes were 1.4m from the ground.
    Code:
                                  o C
                                * |
                              *   |
                            *     |
        A o               *       | h
          | *           *         |
      1.4 |   *       *           |
          |   θ *   * θ           |
        B o - - - o - - - - - - - o D
             1.2  M      15

    Sarah is AB = 1.4
    The tree is CD = h.
    The mirror is at M\!:\;\;BM = 1.2,\;MD = 15

    Since "angle of reflection equals angle of incidence",
    . . we have: . \angle AMB \,=\,\angle CMD \,=\,\theta.

    Hence: . \Delta CDM \sim \Delta ABM


    Therefore: . \frac{h}{15} \;=\;\frac{1.4}{1.2} \quad\hdots\;\text{etc.}

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  5. #5
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    With only using SSS≈, SAS≈ and AA≈, how can you prove these 2 triangles are similar?
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  6. #6
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    I hope this is ok, the tree is very tall in comparison to Sarah.

    In a right-angled triangle, one of the 3 angles is 90 degrees.
    Hence if 2 others are the same,
    the 3rd and final angles are also the same and are 90^o-\theta
    Attached Thumbnails Attached Thumbnails Similar Triangles help-sarah.jpg  
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