Results 1 to 6 of 6

Math Help - tis question is so confusing.hu can solve?

  1. #1
    Junior Member
    Joined
    Jan 2006
    Posts
    38

    tis question is so confusing.hu can solve?

    wat is tis question toking about ?
    Attached Thumbnails Attached Thumbnails tis question is so confusing.hu can solve?-9.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by xiaoz
    wat is tis question toking about ?
    (a)The arc lamp to the shadow, and the man's head to the shadow form two similar triangles.
    The ratio of height to base for the man's triangle is  \frac{1.8}{2.5}
    The ratio of height to base for the arc-lamp triangle is  \frac{x}{10+2.5}
    Because the triangles are similar, the two ratios are equal...
     \frac{1.8}{2.5}=\frac{x}{10+2.5}

    \frac{1.8}{2.5}=\frac{x}{12.5}

    \frac{1.8}{2.5}\cdot12.5=\frac{x}{1}

    \frac{1.8\cdot12.5}{2.5}=x

    \frac{22.5}{2.5}=x

    9=x

    Which means the height of the arc-lamp triangle is 9m.

    (b) This is an interesting question. Imagine the man on a co-ordinate plane with the arc-lamp in the center. He is on point (10,0) and walks to point (10,20).

    We need to first find the distance between him and the lamp. The distance formula is...
    note:  (x_1,y_1)=(0,0)

     \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=d

     \sqrt{\left(10-0\right)^2+\left(20-0\right)^2}=d

     \sqrt{\left(10\right)^2+\left(20\right)^2}=d

     \sqrt{100+400}=d

     \sqrt{500}=d

     22.3606797749979=d

    Just to make it easy to write, we'll say the distance between the man and the lamp is  \sqrt{500}

    Now we know the heights of both triangles, we know they're similar, so we remake our ratios...

     \frac{9}{\sqrt{500}+x}=\frac{1.8}{x}

    and solve from there.
    Last edited by Quick; June 19th 2006 at 05:40 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by xiaoz
    wat is tis question toking about ?
    i) On your drawing/figure as posted, draw a vertical line from the lamp to the ground.
    Let angle BCA = theta.

    In the smaller right triangle,
    tan(theta) = 1.8/2.5

    In the larger right triangle, let x = height of lamp,
    tan(theta) = x/(10+2.5)

    tan(theta) = tan(theta),
    1.8/2.5 = x/(12.5)
    x = (1.8/2.5)(12.5) = 9m ------------answer.

    -------------------
    ii) If the man walks south, you cannot see it on your posted figure.
    Go above and over the lamp. Look down. Walk the man now 20m due south. You have a new right triangle. The position now of the man from the base of the lamp post is, by Pythagorean Theorem,
    d^2 = (10)^2 +(20)^2
    d^2 = 500
    d = 10sqrt(5) m.

    Now go down to the ground again. Look at the man and the lamp post now. You have two new right triangles again.
    Let y = length of the man's shadow now.

    New smaller right triangle:
    ---hypotenuse = ray of light from the lamp = unknown
    ---vertical leg = man's height = 1.8m
    ---horizontal leg = new shadow = y
    ---angle between hypotenuse and horizontal leg = say, alpha.

    New larger right triangle:
    ---hypotenuse = ray of light from the lamp = unknown
    ---vertical leg = lamp's height = 9m
    ---horizontal leg = (10sqrt(5) +y)m
    ---angle between hypotenuse and horizontal leg = alpha also.

    So, in smaller triangle,
    tan(alpha) = 1.8 /y

    In larger triangle,
    tan(alpha) = 9 /(10sqrt(5) +y)

    tan(alpha) = tan(alpha)
    1.8 /y = 9 /(10sqrt(5) +y)
    Cross multiply,
    1.8(10sqrt(5) +y) = 9y
    Divide both sides by 9,
    2sqrt(5) +0.2y = y
    2sqrt(5) = 0.8y
    y = 2sqrt(5) /(0.8)
    y = 2.5sqrt(5) = 5.59 m -------man's shadow now, answer.
    Last edited by ticbol; June 19th 2006 at 11:08 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2006
    Posts
    38

    Smile

    omg.. u all so clever.. but it is confusing ya...
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    W/ this equation
    <br />
\frac{9}{\sqrt{500}+x}=\frac{1.8}{x}<br />
    i have a question sir Quick
    where is the x in the drawing?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Quick's Avatar
    Joined
    May 2006
    From
    New England
    Posts
    1,024
    Quote Originally Posted by ^_^Engineer_Adam^_^
    W/ this equation
    <br />
\frac{9}{\sqrt{500}+x}=\frac{1.8}{x}<br />
    i have a question sir Quick
    where is the x in the drawing?
    x is the distance from the man's feet to the tip of the shadow (it's not the same x as the 1st problem)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 15th 2011, 07:25 PM
  2. Confusing Question
    Posted in the Algebra Forum
    Replies: 4
    Last Post: January 30th 2009, 06:07 PM
  3. Another confusing stats question
    Posted in the Statistics Forum
    Replies: 1
    Last Post: January 30th 2009, 09:51 AM
  4. Confusing question
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: October 2nd 2008, 03:01 AM
  5. Confusing Question
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: October 4th 2006, 04:42 PM

Search Tags


/mathhelpforum @mathhelpforum