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Math Help - Plz solve this trignometry question.Tomorrow is my exam.

  1. #1
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    Plz solve this trignometry question.Tomorrow is my exam.

    Tomorrow is my maths 1 exam and next day maths 2 exam so plz solve these question whole as soon as possible.

    1)As observed from the top of a 100 m tall light house the angles of depression of two ships approaching it are 30^o and 45^o. If one ship is directly behind the other, find the distance between the two sips.

    2)The angles of elevation of the top of a rock at top and foot of a 100 m high tower are 30^o and 45^o respectively>Find the height of the rock.
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  2. #2
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    Hello, prashantvrm!

    They used "nice" angles in these problems.
    We can solve them without a calculator.
    . . I'll walk through the first one.


    1) As observed from the top of a 100 m lighthouse,
    the angles of depression of two ships approaching it are 30 and 45.
    If one ship is directly behind the other, find the distance between the two ships.
    Code:
        A * - - - - - - - - - - - C
          | * * 30
          |   *   *
          |     *     *
      100 |       *       *
          |         *         *
          |       45 *       30 *
          * - - - - - - * - - - - - - *
          B      y      P      x      Q
    The lighthouse is AB = 100.
    The ships are at P and Q.

    \angle CAQ = 30^o, hence: \angle AQB = 30^o
    \angle CAP = 45^o, hence: \angle APB = 45^o

    Let x = PQ,\;y = BP


    In \Delta ABP\!:\;\tan45^o \:=\:\frac{y}{100} \quad\Rightarrow\quad\frac{y}{100} \:=\:1 \quad\Rightarrow\quad y \:=\:100 .[1]

    In \Delta ABQ\!:\;\;\tan30^o \:=\:\frac{100}{x+y} \quad\Rightarrow \quad \frac{1}{\sqrt{3}} \:=\:\frac{100}{x+y} \quad\Rightarrow\quad x+y \:=\:100\sqrt{3} .[2]

    Substitute [1] into [2]: . x + 100 \:=\:100\sqrt{3}


    . . Therefore: . x \;=\;100\sqrt{3} - 100 \;=\;100(\sqrt{3}-1)

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