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Math Help - Triangle

  1. #1
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    Triangle

    need quick help! i dont understand these at all and i have a test tomorrow. damn summer school. anyways
    A tourist walks n km at 4 km/h and then travels 2n km at 36 km/h by bus. Express the total traveling time t in hours as a function of n.
    Answer is t(n) = 11n/36 i dont know how to get it though

    A light 3m above the ground causes a boy 1.8m tall to cast a shadow s meteres long measured along the ground. express s as a function of d, the boy's distance in meters from the light. s(d) = 1.5d is the answer
    Express the area A of a 30-60-90 right triangle as a function of the length of hypotenuse, h. answer is a(h)=h2root3/8
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    is up to his old tricks again! Jhevon's Avatar
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    [quote=PrecalcKid114;59688]need quick help! i dont understand these at all and i have a test tomorrow. damn summer school. anyways
    A tourist walks n km at 4 km/h and then travels 2n km at 36 km/h by bus. Express the total traveling time t in hours as a function of n.
    Answer is t(n) = 11n/36 i dont know how to get it though[\quote]

    recall that \mbox { Speed } = \frac { \mbox { Distance }}{ \mbox { Time }}

    \Rightarrow \mbox { Time } = \frac { \mbox { Distance }}{ \mbox { Speed }}

    Let t_1 be the time the journey on foot takes
    Let t_2 be the time the journey by bus takes
    Let T be the total time

    Thus we have T = t_1 + t_2

    For t_1:

    t_1 = \frac {d}{s} = \frac {n}{4}

    For t_2:

    t_2 = \frac {d}{s} = \frac {2n}{36} = \frac {n}{18}

    Thus we have: T = \frac {n}{4} + \frac {n}{18} = \frac {11n}{36}
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by PrecalcKid114 View Post
    A light 3m above the ground causes a boy 1.8m tall to cast a shadow s meteres long measured along the ground. express s as a function of d, the boy's distance in meters from the light. s(d) = 1.5d is the answer
    did you draw a diagram? we have similar triangles here.


    Let h_1 be the height of the large triangle
    Let b_1 be the base of the large triangle
    Let h_2 be the height of the small triangle
    Let b_2 be the base of the small triangle

    since the triangles are similar, we have:

    \frac {h_1}{b_1} = \frac {h_2}{b_2}

    \Rightarrow \frac {s + d}{3} = \frac {s}{1.8}

    \Rightarrow s + d = 1.6666667s

    \Rightarrow d = 0.6666667s

    \Rightarrow \boxed { s = 1.5d }

    i rounded off a bit, 1.666666.... and 0.66666.... repeats forever
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by PrecalcKid114 View Post
    Express the area A of a 30-60-90 right triangle as a function of the length of hypotenuse, h. answer is a(h)=h2root3/8
    see the diagram below.


    Let A be the area of the triangle
    Let a be the base of the triangle
    Let b be the height of the triangle
    Let h be the hypotenuse of the triangle

    We will use the sine trig ratio

    Since \sin 30 = \frac {b}{h} we have b = h \sin 30 = \frac {h}{2}

    Since \sin 60 = \frac {a}{h} we have a = h \sin 60 = \frac {h \sqrt {3}}{2}

    The area of a triangle is \frac {1}{2} base \times height

    so we have: A = \frac {1}{2} \cdot \frac {h}{2} \cdot \frac {h \sqrt {3}}{2}

    \Rightarrow \boxed { A = \frac {h^2 \sqrt {3}}{8} }

    by the way, we use "^" to show powers, what you typed was h2root3, that is totally different from h^2 root3
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