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A surveyor wishes to find the distance across a swamp. The bearing from A to B (Segment AB is opposite side of triangle) is N 32° W. The surveyor walks 50 meters from A to C, and at the point C the bearing to B is N 68° W. (Segment AC is adjacent side of triangle & Segment BC is hypotenuse side of triangle.)

A) Find bearing from A to C?
B) What is the distance from A to B?

  • Trigonometry -

    Confusing description. Where does the swamp come in ?
    After a few tries I got a right-angled triangle with angle A = 90°
    I sketched a vertical line AP .
    From your description, angle BAP = 32°, leaving
    angle PAC = 58°
    I also drew a vertical from C to Q
    so that angle QCB = 68°

    let angle BCA = Ø
    since AP and CQ are parallel
    angle PAC + angle ACQ = 180°
    58 + Ø+68=180
    Ø = 54°

    OK then....

    A) bearing from A to C is N 58° E
    B) tan 54° = AB/50
    AB = 50tan54 = 68.82 m

  • Trigonometry -

    Reiny thanks a lot for the help! I welcome you with a nice smile! Here is the website where I can show how the problems look/appear like...

    Google this:
    4.8 applications and models Comcast

    Click the first result that appears on the web search page.
    Go to page 8 in the PDF document on problem #33.

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