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Thread: Iíve a problem, please help me

  1. #1
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    Iíve a problem, please help me

    Two radar stations A and B are 16 km apart and A is due north to B. A ship is know to be on a bearing of 150˚ from A and 10 km from B. Show that this information gives two positions for the ship, and calculate the distance between these two positions.

    How can I solve this? Please help me.
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  2. #2
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by geton View Post
    Two radar stations A and B are 16 km apart and A is due north to B. A ship is know to be on a bearing of 150˚ from A and 10 km from B. Show that this information gives two positions for the ship, and calculate the distance between these two positions.

    How can I solve this? Please help me.
    Hi geton!

    Take a look at the diagram. As the ship is 150˚ from A, we can see that this is 30˚ off a 180˚ rotation, so we have angle $\displaystyle \angle BAS =30^{\circ}$.

    Now we have a few values, but we need more angles before we can solve for AS.

    Angle ASB (by the sine rule):

    $\displaystyle \frac{\sin{30^{\circ}}}{10}=\frac{\sin{\angle ASB}}{16}$

    $\displaystyle \sin{\angle ASB}=\frac{\frac{1}{2}}{10}\times 16$

    $\displaystyle \sin{\angle ASB} = \frac{4}{5}$

    $\displaystyle \angle ASB = \sin^{-1}{\left(\frac{4}{5}\right)}$

    There are many values of $\displaystyle \sin^{-1}{\left(\frac{4}{5}\right)}$, the reference angle for which is $\displaystyle 53.13^{\circ}$. To be exact, infinite, because the function is periodic. The solutions we want are:

    $\displaystyle \angle ASB=53.13^{\circ}$ or $\displaystyle \angle ASB=180-53.13=126.87$

    Because those are the quadrants where the sine function is positive and are within the reasonable limits for the angles of a triangle.

    Case 1: $\displaystyle \angle ASB = 53.13^{\circ}$

    Then $\displaystyle \angle ABS = 180-53.13-30=96.87^{\circ}$

    Then by the sine rule again,

    $\displaystyle \frac{AS}{\sin{96.87^{\circ}}}=\frac{10}{\sin{30}}$

    $\displaystyle AS=19.86$km.

    Case 2: $\displaystyle \angle ASB = 126.87^{\circ}$

    Then $\displaystyle \angle ABS = 180-126.87-30=23.13^{\circ}$

    $\displaystyle \frac{AS}{\sin{23.13^{\circ}}}=\frac{10}{\sin{30}}$

    $\displaystyle AS=7.86$km.

    So the difference between the answers is 12 km. I think.
    Attached Thumbnails Attached Thumbnails -ship.jpg  
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  3. #3
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    Thank you so much (DivideBy0).
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