Results 1 to 4 of 4

Math Help - Angles, Degrees, and Arcs

  1. #1
    Newbie
    Joined
    Jul 2006
    Posts
    5

    Angles, Degrees, and Arcs

    Hey there, I've got a few questions.

    I have no idea how to figure out these kind of word problems. I'm taking an online class and I've got no source of help but here. So, is there some sort of formula I must follow with these? Draw a diagram? What does the "bearing" of 42.26 do for me?

    A fire is sighted due west of lookout A. The bearing of the fire from lookout B, 8.6 miles due south of A, is N 4226'W. How far is the fire from B (to the nearest tenth of a mile)?

    Oh, and this one is probably easy, just simple algebra and such, but I dont know what to do... sin A = cos 3A

    Thanks in advance, Juan.
    Last edited by juangohan9; August 7th 2006 at 06:24 AM. Reason: forgot to add something
    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 juangohan9
    Hey there, I've got a few questions.

    I have no idea how to figure out these kind of word problems. I'm taking an online class and I've got no source of help but here. So, is there some sort of formula I must follow with these? Draw a diagram? What does the "bearing" of 42.26 do for me?

    A fire is sighted due west of lookout A. The bearing of the fire from lookout B, 8.6 miles due south of A, is N 4226'W. How far is the fire from B (to the nearest tenth of a mile)?
    42^{\circ}26'=42^{\circ}+\frac{26}{60}^{\circ}=<br />
\frac{<br />
1260+13}{30}

    a right triangle is formed when you draw a line from A to B to the fire, the distance from the Fire to base B is the hypotenuse, the distances from A to B and A to the Fire are the bases of the right triangle. do you know secant?

    \sec\left(\frac{1273}{30}\right)=\frac{\overline{B  F}}{\overline{AB}}

    therefore:

    \sec\left(\frac{1273}{30}\right)=\frac{\overline{B  F}}{8.6}

    thus:

    8.6\sec\left(\frac{1273}{30}\right)=\overline{BF}

    11.7\approx\overline{BF}

    so the distance is 11.7 miles.

    note: \sec A=\cos^{-1}A=\frac{hypotenuse}{adjacent}

    there is also another way of finding it, using tan and the pythagoreum theorum, would you like to see it?
    Attached Thumbnails Attached Thumbnails Angles, Degrees, and Arcs-fire.jpg  
    Last edited by Quick; August 7th 2006 at 07:38 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by juangohan9
    sin A = cos 3A

    Thanks in advance, Juan.
    Thus,
    \sin x - \cos 3x=0
    Use co-function identity,
    \sin x - \sin\left( \frac{\pi}{2} - 3x \right)=0
    ---
    For reference use formula,
    \sin x-\sin y=2\cos \frac{x+y}{2} \sin \frac{x-y}{2}
    ---
    Thus,
    2\cos \left(\frac{\pi}{4}-x\right)\sin \left( \frac{\pi}{4} +2x \right) =0
    Divide by two, thus,
    \cos \left(\frac{\pi}{4}-x\right)\sin \left( \frac{\pi}{4} +2x \right) =0
    Set each factor equal to zero thus,
    \left\{ \begin{array}{c} \cos \left(\frac{\pi}{4}-x\right)=0\\\sin \left( \frac{\pi}{4} +2x \right) =0
    From the first one you have,
    \frac{\pi}{4}-x=\frac{\pi}{2}+\pi k
    Thus,
    x=-\frac{\pi}{4}-\pi k
    To make your answer look cooler sine -k is an integer too is does not matter if you write,
    x=-\frac{\pi}{4}+\pi k
    Since the number being constantly added and subtracted is \pi it does not matter if you write,
    x=-\frac{\pi}{4}+\pi+\pi k
    Simplify,
    x=\frac{3\pi}{4}+\pi k

    For the second one you have,
    \sin \left(\frac{\pi}{4}+2x \right)=0
    Thus,
    \frac{\pi}{4}+2x=\pi k
    Thus,
    2x=-\frac{\pi}{4}+\pi k
    Since the number being constantly subtracted and added is \pi it does not matter if,
    2x=\pi-\frac{\pi}{4}+\pi k
    Thus, simplify,
    2x=\frac{3\pi}{4}+\pi k
    Thus,
    x=\frac{3\pi}{8}+\frac{\pi}{2} k
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,685
    Thanks
    616
    Hello, Juan!

    I'll explain N\,42^o26'\,W
    . . The N means we start by facing North.
    . . Then 42^o26'\,W means we turn 42^o26' to the west.
    And that is the direction indicated by N\,42^o26'\,W.

    Similarly, S30^oE means: face South, turn 30 to the east.


    A fire is sighted due west of lookout A.
    The bearing of the fire from lookout B, 8.6 miles due south of A, is N\,42^o26'\,W
    How far is the fire from B to the nearest tenth of a mile?
    Code:
        F * - - - - - - - * A
            \             |
              \           |
                \         |
                d \       | 8.6
                    \     |
                      \ θ |
                        \ |
                          * B

    The fire is at F. .Angle \theta \,= \,42^o26' \,\approx \,42.4333^o. .Let d = BF.

    In right triangle FAB, we have: . \cos\theta\,=\,\frac{8.6}{d}\quad\Rightarrow\quad d\,=\,\frac{8.6}{\cos\theta}

    Therefore: . d \:= \:\frac{8.6}{\cos42.4333^o} \:= \:11.6521...\:\approx\:11.7 miles.




    Solve for A:\;\;\sin A \:= \:\cos 3A

    Here's a back-door approach to this problem . . .

    If \sin A \,= \,\cos B, then A and B are complementary: . A + B \,=\,90^o

    Look at this right triangle.
    Code:
                      * B
               c   *  *
                *     * a
             *        *
        A *  *  *  *  * C
                b

    We have: . \sin A = \frac{a}{c} .and . \cos B = \frac{a}{c}

    See? . . . Angles A and B are from the same right triangle.

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    Back to the problem: . \sin A \:= \:\cos3A

    Since A and 3A are complementary: . A + 3A \:= \:90^o\quad\Rightarrow\quad 4A \:=\:90^o

    Therefore: . \boxed{A\:=\:22.5^o}


    Note: TPHacker's solution is complete.

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Angles and Arcs
    Posted in the Geometry Forum
    Replies: 8
    Last Post: July 15th 2009, 04:50 AM
  2. Arcs and Angles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: May 17th 2009, 09:36 AM
  3. angles degrees and seconds
    Posted in the Trigonometry Forum
    Replies: 2
    Last Post: May 31st 2008, 07:08 PM
  4. Angles, chords and arcs
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 16th 2008, 08:46 PM
  5. Angles and Degrees!
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 11th 2007, 09:39 PM

Search Tags


/mathhelpforum @mathhelpforum