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Math Help - How high and how far problem?

  1. #1
    Newbie Smokinoakum's Avatar
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    How high and how far problem?

    Hi all... could someone check me here. I have the right process I think, but I'm not 100% sure.

    Here is the problem...

    A student stands a certain distance from a building and measures the angle to the top of the building as 18.8°. The student then walks 24 ft straight towards the building and measures the angle to the top of the building as 20.2°.
    A What is the height of the building, and
    b) How close was the student to the bottom of the building after she had walked the 24 ft?

    This is what I came up with to solve the height of the building.

    A = 159.8
    B = 18.8
    C = 1.4
    a = 339.19
    b = 316.57
    c = 24

    h = 339.19sin(18.8) = 109.3

    This is what I have for the distance from the building.

    A = 20.2
    B = 90
    C = 69.8
    a = 109.3 (the height from the previous answer)
    b = 316.57
    c = 297.07

    thus...She is 297.07ft away.

    How is my logic?

    By the way, I am so glad to be a part of this forum. I am taking Math 122 online and need all the help I can get. I hope I can also contribute to this forum for others as well.

    Neil
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Smokinoakum View Post
    Hi all... could someone check me here. I have the right process I think, but I'm not 100% sure.

    Here is the problem...

    A student stands a certain distance from a building and measures the angle to the top of the building as 18.8°. The student then walks 24 ft straight towards the building and measures the angle to the top of the building as 20.2°.
    A What is the height of the building, and
    b) How close was the student to the bottom of the building after she had walked the 24 ft?

    This is what I came up with to solve the height of the building.

    A = 159.8
    B = 18.8
    C = 1.4
    a = 339.19
    b = 316.57
    c = 24

    h = 339.19sin(18.8) = 109.3

    This is what I have for the distance from the building.

    A = 20.2
    B = 90
    C = 69.8
    a = 109.3 (the height from the previous answer)
    b = 316.57
    c = 297.07

    thus...She is 297.07ft away.

    How is my logic?

    By the way, I am so glad to be a part of this forum. I am taking Math 122 online and need all the help I can get. I hope I can also contribute to this forum for others as well.

    Neil
    You are going to need to provide us with a diagram or some explanation of what A, B, C, a, b, and c are.

    -Dan
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  3. #3
    Newbie Smokinoakum's Avatar
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    Quote Originally Posted by Smokinoakum View Post
    Hi all... could someone check me here. I have the right process I think, but I'm not 100% sure.

    Here is the problem...

    A student stands a certain distance from a building and measures the angle to the top of the building as 18.8°. The student then walks 24 ft straight towards the building and measures the angle to the top of the building as 20.2°.
    A What is the height of the building, and
    b) How close was the student to the bottom of the building after she had walked the 24 ft?

    This is what I came up with to solve the height of the building.

    A = 159.8
    B = 18.8
    C = 1.4
    a = 339.19
    b = 316.57
    c = 24

    h = 339.19sin(18.8) = 109.3

    This is what I have for the distance from the building.

    A = 20.2
    B = 90
    C = 69.8
    a = 109.3 (the height from the previous answer)
    b = 316.57
    c = 297.07

    thus...She is 297.07ft away.

    How is my logic?

    By the way, I am so glad to be a part of this forum. I am taking Math 122 online and need all the help I can get. I hope I can also contribute to this forum for others as well.

    Neil
    The way I got the first measurement for the height was as follows.
    The first angle was given to be 18.8°. I labeled that angle B. Then we're given the base as 24ft. I labeled that c. Then we're told the next angle is 20.2°. I subtracted that from 180 to get the opposite angle, which is 159.8 and I labeled that A. I solved for angle C by adding 159.8 + 18.8 = 1.4°. I then solved for side a and b with the given angles.

    Now for the distance problem I simply took the second angle given as 20.2° and the 90° angle that is next to the building and subtracted that from 180 as well given me 180 - 20.0 - 90 = 69.8. I also used the height of the building of 109.3 as side a. Then I solved for side b and c.

    I hope this clarifies this a bit. I would like to show a diagram but I'm not sure how to do that here. Maybe you could give me a tip for that as well.

    Thanks for you help!

    Neil
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  4. #4
    MHF Contributor
    skeeter's Avatar
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    using "paint" to sketch a diagram ... save as a jpg file and upload as an attachment.

    h = x \tan(20.2)

    h = (x+24)\tan(18.8)

    solve the system for x and h
    Attached Thumbnails Attached Thumbnails How high and how far problem?-bldgheight.jpg  
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  5. #5
    Super Member

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    Lexington, MA (USA)
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    Hello, Smokinoakum!

    A student stands a certain distance from a building
    and measures the angle to the top of the building as 18.8°.
    The student then walks 24 ft straight towards the building
    and measures the angle to the top of the building as 20.2°.

    (a) What is the height of the building?

    (b) How close was the student to the bottom of the building
    after she had walked the 24 ft?


    This is what I came up with to solve the height of the building.

    . . \begin{array}{cccccccc}A&=&159.8^o && a &=& 339.19 \\ B &=& 18.8^o && b &=& 316.57 \\ C &=& 1.4^o && c &=& 24 \end{array}

    h \:=\: 339.19\sin18.8^o \:=\: 109.3


    This is what I have for the distance from the building.

    . . \begin{array}{ccccccccc}A &=& 20.2^o && a &=& 109.3 \\<br />
B &=& 90^o && b &=& 316.37 \\ C &=& 69.8^o && c &=& 297.07 \end{array}

    Thus, she is 297.07ft away.

    How is my logic?

    Once I figured out your labeling scheme, it all became clear.
    Your logic and your work is excellent!


    skeeter's approach is the best way.

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  6. #6
    Newbie Smokinoakum's Avatar
    Joined
    Mar 2011
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    Thanks guys...It's nice to know I'm kinda getting this stuff.

    Thanks for the tip about using paint to label my diagram out.

    Neil
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