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Math Help - Angle Of Depression Question

  1. #1
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    Angle Of Depression Question

    Hi. I have recently attempted to solve the following problem:

    From the top of a 135 foot observation tower, a park ranger sights two forest fires on opposite sides of the tower. If their angles of depression are 42.5 degrees and 32.6 degrees, how far apart are the forest fires?

    Here is how I solved this problem. I drew my diagram, with the 135 foot tower, x's marking each forest fire, and lines from the tower to each x, one with the angle being formed labeled as 42.5 degrees, the other as 32.6 degrees. I labeled the distance between Fire #1 and the tower a, and the distance between fire number 2 and the tower b. Then I solved for a and b. To solve for a, I wrote Tangent 32.6 = a/135. I think this is correct because a is opposite the angle, and the tower is adjacent to it. Then I multiplied 135 by the tangent of 32.6 (which is about .64), and got about 86.3 for a. Then I solved for b- Tangent 42.5 = b/135, solved for b the same way, and got 123.7 (Tan 42.5 =.92*135 = 123.7). The distance between the two fires, is a+b, so I added 123.7 to 86.3 (not these rounded figures, the actual answers the calculators gave me), and I got an answer of about 210.

    However, the answer key I consulted with later said the correct distance between the fires was 358 feet. Can someone please explain to me where I went wrong, as I was pretty confident with my answer? Thank you so much in advance.
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  2. #2
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    e^(i*pi)'s Avatar
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    Well using the sine rule I get an answer of 210m like you

    Let the observation tower be angle C and the fires at 32.6 degrees and 42.5 degrees be A and B respectively. Drawing a perpendicular line down from C we get the height of the tower known to be 135m and two triangles. Call the point where this perpendicular line intersects the line AB as point D.

    Thus we have two triangles: CAD and CBD. Since a triangle's angles always add to 180 degrees then it's easy to find the other two angles to be 57.4 and 47.5 respectively.

    Sine rule: \dfrac{a}{\sin A} = \dfrac{b}{\sin B} - this uses the convention that side a is opposite angle A and so on.


    For triangle CAD:

    \dfrac{AD}{\sin(32.6)} = \dfrac{135}{\sin(57.4)} \: \rightarrow \: AD = \dfrac{135 \sin(32.6)}{\sin(57.4)} (135=CD = the height of the tower)

    (I am leaving my answers in trig form to avoid rounding errors at this stage)

    For triangle CBD:

    \dfrac{BD}{\sin(42.5)} = \dfrac{135}{\sin(47.5)} \: \rightarrow \: BD = \dfrac{135 \sin(42.5)}{\sin(47.5)}


    The length is given by AB = AD + BD = \dfrac{135 \sin(32.6)}{\sin(57.4)} + \dfrac{135 \sin(42.5)}{\sin(47.5)} \approx 210

    See attached for diagram (I scanned it, hence the very unimaginative title): Attachment 18891
    Last edited by e^(i*pi); September 11th 2010 at 01:49 PM. Reason: correcting previously incorrect fractions
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  3. #3
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    Hello, geoff9999!

    You used the wrong angles . . .


    \text{From the top of a 135-foot tower, a forest ranger sights two forest fires}
    \text{on opposite sides of the tower. }\,\text{ If their angles of depression are }42.5^o
    \text{ and }32.6^o,\,\text{ how far apart are the forest fires?}

    In the diagram, "d" represents "degrees".


    Code:
                            P
          W - - - - - - - - * - - - - - - - - - E
                    42.5d * | * 32.6d
                        *   |   *
                      *47.5d|57.4d*
                    *       |       *
                  *         |135      *
                *           |           *
              *             |             *
            *               |               *
        Q *-----------------*-----------------* R
          : - - -  a  - - - S - - -  b  - - - :

    The tower is: PS = 135 ft.

    One fire is at Q. .Let a \,=\,QS.
    \angle WPQ = 42.5^o \quad\Rightarrow\quad \angle SPQ = 47.5^o

    The other fire is at R. .Let b \,=\,SR.
    \angle EPR = 32.6^o \quad\Rightarrow\quad \angle SPR = 57.4^o


    In right triangle PSQ\!:\;\tan47.5^o \,=\,\dfrac{a}{135} \quad\Rightarrow\quad a \,=\,135\tan47.5^o \,\approx\,147\text{ ft}

    In right triangle PSR\!:\;\tan57.4^o \,=\,\dfrac{b}{135} \quad\Rightarrow\quad b \,=\,135\tan57.4^o \,\approx\,211\text{ ft}


    Therefore: . QR \:=\:a + b \:=\: 147 + 211 \:=\:358\text{ ft}
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