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Math Help - Height of a Tower

  1. #1
    Member purplec16's Avatar
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    Height of a Tower

    From a point P on level ground, the angle of elevation of he top of the tower is 26.83 degrees. From a point 25.0 meters closer to the tower and on the same line with P and the base of the tower, the angle of elevation of the top is 53.5 degrees. Approximate the height of the tower.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by purplec16 View Post
    From a point P on level ground, the angle of elevation of he top of the tower is 26.83 degrees. From a point 25.0 meters closer to the tower and on the same line with P and the base of the tower, the angle of elevation of the top is 53.5 degrees. Approximate the height of the tower.
    See the attached diagram. Note that triangles PTB and ATB are both right triangles

    Do you think you can figure out how to get to the answer?
    Attached Thumbnails Attached Thumbnails Height of a Tower-tower.jpg  
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    Member purplec16's Avatar
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    I know how to draw it I jus dont know how to work it
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by purplec16 View Post
    I know how to draw it I jus dont know how to work it
    Use the big right triangle to solve for P (your answer will be in terms of h and require the use of trig functions). In a similar manner, use the small triangle to solve for P - 25. Your answer will also have h in it. Solve for P again there. Then, equate the two expressions you got for P and solve the resulting equation for h.

    Think you can manage that? Try
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  5. #5
    Member purplec16's Avatar
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    I tried, I didn't get it
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by purplec16 View Post
    I tried, I didn't get it
    Post what you've tried so we can see where you're having trouble, and fix it.
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  7. #7
    Member purplec16's Avatar
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    Ok u said to first solve for P so...
    I did sin 53.5= h/ P-25
    25 sin 53.5 = h
    P=45.09

    Then for the smaller triangle h would be
     20 sin 26.83 = 9.027

    You said something bout equate them, but the answer in the book is 20.2m, so
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by purplec16 View Post
    Ok u said to first solve for P so...
    I did sin 53.5= h/ P-25
    25 sin 53.5 = h
    P=45.09

    Then for the smaller triangle h would be
     20 sin 26.83 = 9.027

    You said something bout equate them, but the answer in the book is 20.2m, so
    ok, first of all, sine does not come into play here. you have the opposite and adjacent sides!

    secondly, how did you find out that P = 45.09 from the first triangle only?! you have one equation and two unknowns. There is no way you can solve for P like that! the answer has to have h in it.

    make another attempt
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  9. #9
    Member purplec16's Avatar
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    tan 53.5 = h/25

    25 tan 53.5= h

    h= 33.79 m


    tan 26.83 = h/P -25

    25 tan 26.83= h

    h= 12.64 m

    P= 21.15

    So how do you find h?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by purplec16 View Post
    tan 53.5 = h/25

    25 tan 53.5= h

    h= 33.79 m
    how can you find h from what you have here? that 25 should not be there. the base of the small triangle is P - 25

    tan 26.83 = h/P -25

    25 tan 26.83= h

    h= 12.64 m

    P= 21.15

    So how do you find h?
    i do not know what happened here. first, your -25 in the first line just disappeared into thin air. then, you assume that P = 25, which it doesn't.
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  11. #11
    Member purplec16's Avatar
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    Well I really dont understand what to do so can you just show me now please I tried twice
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    Quote Originally Posted by purplec16 View Post
    Well I really dont understand what to do so can you just show me now please I tried twice
    From the big triangle,

    \tan 26.83 = \frac hP

    \Rightarrow P = \frac h{\tan 26.83} ...............(1)

    From the small triangle,

    \tan 53.5 = \frac h{P - 25}

    \Rightarrow P = \frac h{\tan 53.5} + 25 .............(2)

    Now equate (1) and (2)

    \Rightarrow \frac h{\tan 26.83} = \frac h{\tan 53.5} + 25

    now solve for h
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  13. #13
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    hi purplec,

    There are 3 sides to a right-angled triangle and 3 angles.
    One angle is 90 degrees and the other two are acute.

    Sin, Cos and Tan give the ratio of 2 sides with respect to one of the acute angles.

    You need to locate the sides "opposite to" and "adjacent to" an angle.
    The longest side is the hypotenuse.

    SinA=\frac{opposite}{hypotenuse}

    CosA=\frac{adjacent}{hypotenuse}

    TanA=\frac{opposite}{adjacent}

    You can use SOHCAHTOA to help remember until you master it.

    The small triangle is inside the big one.

    The height "h" is the opposite to both angles and P is on the adjacent.

    We use Tan if we are dealing with opposite and adjacent.

    Big Triangle

    Tan(26.83^o)=\frac{h}{P}

    multiply both by P

    [P]Tan(26.83^o)=\frac{hP}{P}=h

    divide by the Tan term for P

    \frac{PTan(26.83^o)}{Tan(26.83^o)}=P=\frac{h}{Tan(  26.83^o)}


    Small Triangle

    Tan(53.5)=\frac{h}{(P-25)}

    multiply both sides by (P-25)

    (P-25)Tan(53.5^o)=\frac{(P-25)h}{(P-25)}=h

    Multiply out the left

    PTan(53.5^o)-25Tan(53.5^o)=h

    Bring the P term and h together by subtracting h from both sides

    PTan(53.5^o)-25Tan(53.5^o)-h=h-h=0

    add the 25Tan term to both sides

    PTan(53.5^o)-25Tan(53.5^o)+25Tan(53.5^o)-h=25Tan(53.5^o)

    PTan(53.5^0)-h=25Tan(53.5^o)


    Use your result for h from the big triangle

    P=\frac{h}{Tan(26.83^o)}

    Therefore

    \frac{hTan(53.5^o)}{Tan(26.83^o)}-h=25Tan(53.5^o)

    multiply both sides by Tan(26.83^o)

    hTan(53.5^o)-hTan(26.83^o)=25Tan(53.5^o)Tan(26.83^o)

    Write this using h as a factor

    h[Tan(53.5^o)-Tan(26.83^o)]=25Tan(53.5^o)Tan(26.83^o)

    Divide both sides by the factor of h

    h=\frac{25Tan(53.5^o)Tan(26.83^o)}{Tan(53.5^o)-Tan(26.83^o)}

    It's of course simplest to use your calculator to find the two values of Tan.
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  14. #14
    Member purplec16's Avatar
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    But I dont see how you transition in the big triangle and little triangle all into one formula
    Use your result for h from the big triangle
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  15. #15
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    Hi purplec,

    The big triangle gave us

    P=\frac{h}{Tan(26.83^o)}

    This does not allow us to find h unless we know P.
    Or... to find P, we need to know h.

    Remember how simultaneous equations work?

    If there are 2 things you do not know, then you need a 2nd clue.
    For instance x+2=5, then you know x is 3 or you can work it out.
    That's one clue needed to find one unknown.

    But if we had x+y =5, we are stuck.
    They could be 0 and 5, or 1 and 4, or 2 and 3, or -2 and 7 and so on.

    When we have 2 unknowns, we need 2 clues.
    In simultaneous equations, you may have just cancelled terms as follows.

    x+y=5
    x-y=1

    add them... 2x+y-y=5+1=6 gets rid of y, so 2x=6, so x=3 and then y=2

    Or.... we could write x=1+y from the 2nd equation,

    then use this in the other

    (1+y)+y=5, so 2y+1=5, so 2y=5-1=4, so y=2 and x=3

    Do you see the idea ?
    You may have covered these in class.

    We use the same type of trick here.

    H is the vertical height of both triangles.

    The bases are different, one is P, the other is P-25

    However we can use Tan(angle), using both angles given to write
    out our two clues, since the sides we are dealing with involve "opposite"
    and "adjacent".

    We write P "in terms of h" using one triangle,
    just as we did with the x and y values above.

    Then we swop that into the 2nd equation, effectively getting rid of P
    just as we got rid of x above to figure out what y is.

    We got rid of P (wrote it in terms of h) because the question asked us to find h.

    Study this idea, because it's an important maths technique.

    You are only getting used to trigonometry,
    so this problem contains quite a bit of calculations.

    But it's important that you master the idea of cancelling one of 2 unknowns
    if you only want one of them.
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