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Mathematics LibreTexts

13.2: Recommendation

  • Page ID
    14045
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    Recommendation

    Where possible draw a sketch of the figure(s) involved in the problem.
    Some of the problems here are simple. The solution can be worked out fast by quick reasoning. The benefit of these problems is not to find the solution by reasoning, but by learning algebraic steps applicable to more challenging situations. Solving application problems is a skill. Cultivate it.
    My approach is not to read a problem first until I understand it. Many of my colleagues will disagree with me. I propose creating a preamble in which you write a symbol or set of symbols for each element in the problem. (Refer to exercise 10.2.9).
    Look at the end of the problem where you usually find what to solve for. \(x\) is the number looked for. Write this as the first step in the preamble (list of symbols in non-trivial exercises whose solution we seek).
    Next start reading the problem from the beginning and develop the preamble step by step. Write the symbol(s) for each step on a new line.
    Finish the preamble by converting all problem steps into symbols. Now look at your preamble and gain an overview of the your problem.
    It should be easier to obtain an equation using preamble symbols. Then solve the equation by the method introduced this far for linear equations. (A linear equation has a variable to the first degree (exponent).)

    Example \(\PageIndex{1}\)

    The perimeter of a rectangle is \(68\) ft. Its length is \(7\) more than twice its width. Find the dimensions of the rectangle.

    Solution:

    Preamble:

    \(x\) is the width (see the diagram).
    Length: \(7\) more than twice width: \(2x+7\).

    (0,0)(-25,-2) (0,0)(1,0)10 (0,0)(0,1)5 (10,5)(-1,0)10 (10,5)(0,-1)5 (9.0,2.5)\(x\) (3.5,0.5)\(2x+7\)

    \(\)
    Equation:

    \(\begin{array}{rcl lll} \hbox{The perimeter is }\ \ P&=&2(L+W)\\[5pt] 68&=&2(2x+7+x)\\[5pt] 68&=&2(3x+7)\\[5pt] 68&=&6x+14\\[5pt] 68-14&=&6x+14-14\\[5pt] 54&=&6x\\[10pt] \displaystyle \frac{54}{6}&=&\displaystyle \frac{6}{6}x\\[10pt] 9&=&x \end{array}\)

    (0,0)(-25,-2) (0,0)(1,0)12 (0,0)(0,1)5 (12,5)(-1,0)12 (12,5)(0,-1)5 (11.0,2.5)\(9\) (0.5,0.5)\(2(9)\!+\!7\!=\!18\!+\!7\!=\!25\)

    Example \(\PageIndex{2}\)

    Angle 1 of a triangle is \(11\) more than three times angle 2. Angle 3 is \(15\) less than \(4\) times angle 2. Find the measure of all three angles.

    Then find the complement and the supplement of the largest angle.

    Hint: The sum of the angles of a triangle is \(180^{\circ}\).

    Solution:

    Preamble:

    Let \(x\) be the measure of angle 2.
    Angle 1 of is \(11\) more than three times angle 2: \(3x+11\).
    Angle 3 is \(15\) less than \(4\) times angle 2: \(4x-15\).

    (0,0)(-29,-7) (0,0)(1,0)12.0 (0,0)(1,1)8 (8,8)(1,-2)3.9 (9.0,0.5)\(x\) (1.3,0.5)\(3x+11\) (4.35,3.9)\(4x-15\)

    \(\)

    Equation:

    \(\begin{array}{rcl lll} \hbox{The sum of the angles}&=&\hbox{\)180^\(}\\[4pt] x+(3x+11)+(4x-15)&=&180\\[4pt] x+3x+11+4x-15&=&180\\[4pt] 8x-4&=&180\\[4pt] 8x-4+4&=&180+4\\[4pt] 8x&=&184\\[4pt] \displaystyle \frac{8}{8}x&=&\displaystyle \frac{184}{8}\\[10pt] x&=&23\\[4pt] \end{array}\)

    Angle 2=\(23^{\circ}\)
    Angle 1=\(3(23)+11=69+11=80^{\circ}\)
    Angle 3=\(4(23)-15=92-15=77^{\circ}\)

    (0,3)(-25,-4) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (8.0,0.5)\(23^{\circ}\) (1.3,0.5)\(80^{\circ}\) (5.8,4.9)\(77^{\circ}\)

    \(\)
    The complement of \(80^{\circ}\) is \(90-80=10^{\circ}\).

    The supplement of \(80^{\circ}\) is \(180-80=100^{\circ}\)

    Example \(\PageIndex{3}\)

    Two lines intersect. one of the acute angles is \(45^{\circ}\).

    Find the measures of the other three angles.

    Solution:

    Preamble:

    (0,0)(-27,2) (0,0)(3,1)6 (0,0)(-3,-1)6 (0,0)(3,-1)6 (0,0)(-3,1)6 (0,0.5)\(A\) (-0.4,-1.0)\(B\) (1.5,-0.3)\(C\) (-2.3,-0.3)\(45^{\circ}\)

    \(\)
    Equation:

    The four angles make a complete
    rotation (\(360^{\circ}\)).

    \(A\) added to \(45^{\circ}\) measures a half rotation or \(180^{\circ}\).

    Similarly \(B+45=180^{\circ}\) also. Thus \(B=180-45=135^{\circ}=A\).

    \(B\) is the supplement of \(45^{\circ}\). So is \(A\).

    \(C+45=180^{\circ}\). Thus \(C=180-45=135^{\circ}\). \(C\) is the supplement of \(45^{\circ}\).

    The pair of angles \(45^{\circ}\) and \(C\) are said to be vertical angles.

    The pair of angles \(A\) and \(B\) are also called vertical angles.

    Example \(\PageIndex{4}\)

    Two parallel lines are intersected by a transversal.

    One acute angle measures \(40^{\circ}\).

    (0,0)(-55,-2) (0,0)(1,0)20 (0,0)(-1,0)3 (0,6)(1,0)20 (0,6)(-1,0)3 (0,0)(2,1)18 (0,0)(-2,-1)4 (0,-1.5)\(A\) (1.4,0.3)\(40^{\circ}\) (8.8,5.0)\(B\) (14.0,6.3)\(C\)

    \(\)
    Find (the measures of) angles \(A\), \(B\) and \(C\).

    Solution:

    \(40^{\circ}+A=180^{\circ}\). \(A=140^{\circ}\) is the supplement of \(40^{\circ}\).

    \(40^{\circ}\) \(B\) are alternate interior angles.
    Their measures are equal. Thus \(B=40^{\circ}\).

    \(B\) and \(C\) are vertical angles. These angles are congruent
    (equal in all respects). Thus \(C=40^{\circ}\).

    (0,2)(-55,-15) (0,0)(1,0)20 (0,0)(-1,0)3 (0,6)(1,0)20 (0,6)(-1,0)3 (0,0)(2,1)18 (0,0)(-2,-1)4 (0,-1.5)\(140^{\circ}\) (1.4,0.3)\(40^{\circ}\) (8.8,5.0)\(40^{\circ}\) (14.0,6.3)\(40^{\circ}\)

    \(\)
    The space between the two vertical lines is the interior of the parallel lines. The region above the top line and the region below the bottom line is the exterior.

    A straight angle measures \(180^{\circ}\).

    A right angle \(90^{\circ}\) is called .

    An acute angle measures less than \(90^{\circ}\).

    An obtuse angle measures less than \(90^{\circ}\).

    Example \(\PageIndex{5}\)

    Find the measures of the two angles.

    (0,0)(-16,4) (10,6)(1,0)8 (10,6)(-1,0)3 (11.5,6)(2,1)7 (7.2,6.3)\(2x+47\) (14.0,6.3)\(x-17\)


    Solution:

    The sum of the measures is \(180^{\circ}\).

    \(\begin{array}{rcl lll} (2x+47)+(x-17)&=&180\\[6pt] 2x+47+x-17&=&180\\[6pt] 3x+30&=&180\\[6pt] 3x+30-30&=&180-30\\[6pt] 3x&=&150\\[10pt] \displaystyle \frac{3}{3}x&=&\displaystyle \frac{150}{3}\\[10pt] x&=&50 \end{array}\)

    (0,0)(-16,3) (6,6)(1,0)15

    (11.5,6)(2,1)7 (9.1,9.3)\(2(50)+47\) (7.6,7.9)\(=100+47\) (7.6,6.3)\(=147^{\circ}\)

    (14.0,6.3)\(50-17=33^{\circ}\)

    Example \(\PageIndex{6}\)

    An isosceles triangle has two congruent sides.
    One of these is 2 more than double the non-congruent side.
    The perimeter is \(174\) ft. Find the triangle’s dimensions.

    Solution:

    Preamble:

    (0,0)(-30,-2) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (4.5,-1.0)\(x\) (-0.9,3.5)\(2x+2\) (8.8,3.5)\(2x+2\)
    Equation:

    \(\begin{array}{rcl lll} \hbox{Perimeter}&=&\hbox{Perimeter}\\[5pt] x+(2x+2)+(2x+2)&=&174\\[5pt] x+2x+2+2x+2&=&174\\[5pt] 5x+4&=&174\\[5pt] 5x+4-4&=&174-4\\[5pt] 5x&=&170\\[5pt] \displaystyle \frac{5}{5}x&=&\displaystyle \frac{170}{5}\\[10pt] x&=&34 \end{array}\)

    (0,0)(-25,-6) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (4.5,-1.0)\(34\) (8.8,3.5)\(2(34)+2=68+2\) (8.8,3.5)\(2(34)+2\) (9.2,2.3)\(=68+2=70\) (4.5,3.5)\(70\)

    Example \(\PageIndex{7}\)

    Triangle \(ABC\) is shown on the right. Trapezoid \(ACED\) is inscribed in the triangle. Show that the triangle has a greater parameter than the trapezoid.

    The key is triangle \(BDE\).

    (0,0)(-18,-3) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (3.5,3.5)(1,0)5 (-1.0,-1.0)\(A\) (6.4,7.5)\(B\) (9.8,-1.0)\(C\) (3.5,3.5) (8.7,3.5) (2.0,3.0)\(D\) (9.4,3.0)\(E\)

    \(\)

    (0,5)(-27,-12) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (-1.0,-1.0)\(A\) (6.4,7.5)\(B\) (9.8,-1.0)\(C\)

    \(\)

    (0,0)(-32,-8) (0,0)(1,0)10.2 (3.5,3.5)(-1,-1)3.5 (8.7,3.5)(1,-2)1.7 (3.5,3.5)(1,0)5 (-1.0,-1.0)\(A\) (9.8,-1.0)\(C\) (3.5,3.5) (8.7,3.5) (2.0,3.0)\(D\) (9.4,3.0)\(E\)

    \(\)
    Solution:

    Line segments \(DA\), \(AC\) and \(CE\) are part of both the triangle and the trapezoid.

    Side \(DE\) which completes the trapezoid is shorter than the sum of \(DB\) and \(EB\) which completes the triangle.

    The triangle has the greater perimeter.

    Example \(\PageIndex{1}\)

    Example 8:

    Side 1 of a triangle is \(2\) ft more than side 2. Side 3 is \(2\) ft. less than side 2.

    The side of a square is \(9\) ft. The perimeter of the triangle and the perimeter of square is the same number.

    What are the dimensions of the triangle?

    Solution:

    Preamble:

    (0,5)(-22,0) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (5.0,-1.0)\(x\) (-0.9,3.0)\(x+2\) (9.4,3.0)\(x-2\) (21.5,2.0)\(9\) (18,-1.0)\(9\) (16,0)(1,0)5 (16,0)(0,1)5 (21,5)(-1,0)5 (21,5)(0,-1)5

    \(\)
    Let \(x\) be side 2. Then \(x+2\) and \(x-2\)
    are the other sides of the triangle

    Perimeter of triangle:

    \((x-2)+x+(x+2)=x-2+x+x+2=3x\)
    Perimeter of square: \(4(9)=36\)

    Equation:

    \(\begin{array}{rcl lll} \hbox{Perimeter of triangle}&=&\hbox{Perimeter of square}\\[5pt] 3x&=&36\\[10pt] \displaystyle \frac{3}{3}x&=&\displaystyle \frac{36}{3}\\[10pt] x&=&12 \end{array}\)

    The dimensions of the triangle are
    \(12-2=10\), \(12\) and \(12+2=14\) ft.

    (0,0)(-25,-6) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (5.0,-1.0)\(12\) (0.1,3.0)\(14\) (9.4,3.0)\(10\)

    Example \(\PageIndex{9}\)

    Find the three angles in the figure on the right.

    (0,0)(-30,0) (0,0)(1,0)5.0 (0,0)(-1,0)5 (0,0)(-1,2)2.0 (0,0)(1,1)3.5 (2.5,1.0)\(3x+6\) (-5.1,1.0)\(5x+5\) (-0.4,1.0)\(5x\)


    Solution:

    Preamble:

    The sum of the three angles is a straight angle.

    Equation:

    \(\begin{array}{rcl lll} \hbox{The sum of the three angles}&=&180^{\circ}\\ (5x+5)+5x+(3x+6)&=&180\\[4pt] %5x+5+5x+3x+6&=&180\\[4pt] 13x+5+6&=&180\\[4pt] 13x+11&=&180\\[4pt] 13x+11-11&=&180-11\\[4pt] 13x&=&169\\[4pt] \displaystyle \frac{13}{13}x&=&\displaystyle \frac{169}{13}\\[12pt] x&=&13 \end{array}\)

    \(5x+5=5(13)+5=65+5=70\),

    \(5x=5(13)=65\),

    \(3x+6=3(13)+6=39+6=45\).

    (0,0)(-30,-25) (0,0)(1,0)5.0 (0,0)(-1,0)5 (0,0)(-1,2)2.0 (0,0)(1,1)3.5 (2.5,1.0)\(45^{\circ}\) (-5.1,1.0)\(70^{\circ}\) (-0.4,1.0)\(65^{\circ}\)

    Exercises 13

    1. The perimeter of a rectangular garden is \(104\) ft. Its length is \(8\) ft less than twice its width. Find the dimensions of the rectangle.

    2. Angle \(1\) of a triangle is \(2\) more than four times angle \(2\). Angle \(3\) is \(9\) less than \(6\) times angle \(2\). Find the measure of all three angles.
      Then find the complement and the supplement of the middle angle.

      Hint: The sum of the angles of a triangle is \(180^{\circ}\).

    3. Two rural roads intersect. One of the acute angles formed by the roads is \(53^{\circ}\). Find the measures of the other three angles.

      (0,6)(-18,-4) (0,0)(3,1)6 (0,0)(-3,-1)6 (0,0)(3,-1)6 (0,0)(-3,1)6 (-0.2,0.5)\(C\) (-0.8,-1.0)\(B\) (1.5,-0.3)\(A\) (-3.2,-0.3)\(53^{\circ}\) (3.0,1) (-3.0,1) (3.0,-1) (-3.0,-1) (3.0,1.5)\(1\) (-3.0,1.5)\(2\) (3.0,-2.0)\(3\) (-3.0,-2.0)\(4\)

      \(\)

    4. A pair of railroad tracks (parallel lines) is intersected by a road (transversal). A gate at \(C\) is to swing between the track and the road. Through how many degrees should the gate swing? (What is the measure of \(C\)?)

      (0,11)(-15,-2) (0,0)(1,0)20 (0,0)(-1,0)3 (0,6)(1,0)20 (0,6)(-1,0)3 (0,0)(2,1)18 (0,0)(-2,-1)4 (11.5,6)(-2,-1)12 (11.45,6)(-2,-1)12 (11.4,6)(-2,-1)12 (11.35,6)(-2,-1)12 (11.3,6)(-2,-1)12 (11.25,6)(-2,-1)12 (11.2,6)(-2,-1)12 (11.15,6)(-2,-1)12 (0,-1.5)\(A\) (1.9,0.3)\(80^{\circ}\) (8.2,5.0)\(C\) (14.0,6.3)\(B\)

      \(\)

    5. A bridge crosses a river at an angle. Angular building \(1\) sits on the river’s edge on one side of the bridge. Angular building \(2\) is constructed on the river’s edge on the other side of the bridge. The angle for building \(2\) is twice that of building \(1\). The angles left between the buildings and the bridge are shown in the figure below. Find the measure of the acute angle between the bridge and the river.

      \(\)

      (0,14)(-1,-3) (10,6)(1,0)8 (10,6)(-1,0)7 (10,1)(1,0)8 (10,1)(-1,0)7 (11.5,6)(-2,-1)12 (11.5,6.1)(-2,-1)12 (11.5,6.2)(-2,-1)12 (11.5,6.3)(-2,-1)12 (11.5,6)(2,1)7 (11.5,6.1)(2,1)7 (11.5,6.2)(2,1)7 (11.5,6.3)(2,1)7 (11.5,6)(-2,-1)1 (11.5,6.1)(-2,-1)1 (11.5,6.2)(-2,-1)1 (11.5,6.3)(-2,-1)1 (11.5,6.4)(4,1)8 (11.5,6.0)(-1,2)3 (12.4,2.9)river (1.4,3.0)bridge (16.9,6.9)angular building \(1\) (4.0,9.5)angular (4.0,8.5)building \(2\) (8.0,6.2)\(2x^{\circ}\) (14.4,6.2)\(x^{\circ}\) (17.5,8.2)\(18^{\circ}\) (11.0,8.0)\(36^{\circ}\) (19.0,3)not drawn to scale

      \(\)

    6. A log is in a horizontal position. It is suspended by two cables joined at a point to form an isosceles triangle with the log as one side of the triangle. Each cable is \(5\) ft less than \(3\) times the length of the log. Find the length of one cable if the perimeter is \(95\) ft.

    7. A block of wood has a cross section in the form of an equilateral (all three sides are congruent) triangle \(ABC\) as shown on the right. Each side is \(10\) ft long. Trapezoid \(ADEC\) is formed by cutting off the top half of the triangle. What is the length of the top \(DE\) of the trapezoid? \(\)

      (0,4)(-5,-1) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (3.5,3.5)(1,0)5 (-1.0,-1.0)\(A\) (6.4,7.5)\(B\) (9.8,-1.0)\(C\) (3.5,3.5) (8.7,3.5) (2.0,3.0)\(D\) (9.4,3.0)\(E\) (5.4,-1.0)\(10\) (0.6,1.5)\(5\) (10.2,1.5)\(5\) (4.0,5.5)\(5\) (7.9,5.5)\(5\)

      \(\)

      (0,0)(-19,-4.5) (0,0)(1,0)10.2 (3.5,3.5)(-1,-1)3.5 (8.7,3.5)(1,-2)1.7 (3.5,3.5)(1,0)5 (-1.0,-1.0)\(A\) (9.8,-1.0)\(C\) (3.5,3.5) (8.7,3.5) (2.0,3.0)\(D\) (9.4,3.0)\(E\)

      \(\)

    8. A pile of logs is stacked in the triangular shape shown in the triangle on the right.

      Side \(1\) of a triangle is \(3\) ft more than side \(2\). Side \(3\) is \(3\) ft. less than side \(2\).
      The side of a square is \(15\) ft. The perimeter of the triangle and the square is the same number.
      What are the dimensions of the triangle?

      (0,8)(-6,1) (0,0)(1,0)11.5 (0,0)(2,3)3.6 (11.5,0)(-3,2)8.0 (5.0,-1.0)\(x\) (-2.1,3.0)\(x-3\) (9.4,3.0)\(x+3\) (25.5,2.0)\(15\) (22,-1.0)\(15\) (20,0)(1,0)5 (20,0)(0,1)5 (25,5)(-1,0)5 (25,5)(0,-1)5 (1,0.5)(1,0)10 (1.5,1.5)(1,0)8 (2.0,2.5)(1,0)6 (2.5,3.5)(1,0)4 (3.0,4.5)(1,0)2 (20.5,0.5)(1,0)5 (20.5,1.5)(1,0)5 (20.5,2.5)(1,0)5 (20.5,3.5)(1,0)5 (20.5,4.5)(1,0)5

      \(\)

    9. Two searchlights are programmed to illuminate the sky at angles configured in the picture on the right. Find the three angles. \(\)

      (0,5)(-15,-0) (0,0)(1,0)5.0 (0,0)(-1,0)5 (0,0)(-1,2)3.0 (0,0)(1,1)5.5 (3.5,2.5)\(3x+17\) (-6.6,2.5)\(5x-6\) (-0.8,2.5)\(x+7\) (3.5,7.0)searchlight 1 (-8.5,7.0)searchlight 2

    1. The perimeter of a rectangular garden is \(104\) ft. Its length is \(8\) ft less than twice its width. Find the dimensions of the rectangle.

      Solution:

      Preamble:

      \(x\) is the width.
      \(2x-8\) is the length.

      (0,0)(-25,-1) (0,0)(1,0)10 (0,0)(0,1)5 (10,5)(-1,0)10 (10,5)(0,-1)5 (9.0,2.5)\(x\) (3.5,0.5)\(2x-8\)

      \(\)
      Equation:

      The perimeter is
      \(\begin{array}{rcl lll} P&=&2(L+W)\\[5pt] 104&=&2(2x-8+x)\\[5pt] 104&=&2(3x-8)\\[5pt] 104&=&6x-16\\[5pt] 104+16&=&6x-16+16\\[5pt] 120&=&6x\\[10pt] \displaystyle \frac{120}{6}&=&\displaystyle \frac{6}{6}x\\[10pt] 20&=&x\\[5pt] x&=&20 \end{array}\)

      (0,0)(-25,-2) (0,0)(1,0)13 (0,0)(0,1)5 (13,5)(-1,0)13 (13,5)(0,-1)5 (11.3,3.0)\(20\) (0.5,0.5)\(2(20)\!-\!8\!=\!40\!-\!8\!=\!32\)

      \(\)
      The dimensions of the rectangle are \(32\) by \(20\) ft.

    2. Angle \(1\) of a triangle is \(2\) more than four times angle \(2\).
      Angle \(3\) is \(9\) less than \(6\) times angle \(2\).
      Find the measure of all three angles.
      Then find the complement and the supplement of the middle angle.

      Hint: The sum of the angles of a triangle is \(180^{\circ}\).

      Solution:

      Preamble:

      Let \(x\) be the measure of angle \(2\).
      Angle \(1\) of is \(2\) more than four times angle \(2\): \(4x+2\).
      Angle \(3\) is \(9\) less than \(6\) times angle \(2\): \(6x-9\).

      (0,0)(-24,-8) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (9.0,0.5)\(x\) (1.3,0.5)\(4x+2\) (4.35,3.4)\(6x-9\)

      \(\)
      Equation:

      \(\begin{array}{rcl lll} \hbox{The sum of the angles}&=&\hbox{\)180^\(}\\[4pt] x+(4x+2)+(6x-9)&=&180\\[4pt] x+4x+2+6x-9&=&180\\[4pt] 11x-7&=&180\\[4pt] 11x-7+7&=&180+7\\[4pt] 11x&=&187\\[10pt] \displaystyle \frac{11}{11}x&=&\displaystyle \frac{187}{11}\\[11pt] x&=&17 \end{array}\)

      Angle \(2=17^{\circ}\)
      Angle \(1=4(17)+2=68+2=70^{\circ}\)
      Angle \(3=6(17)-9=102-9=93^{\circ}\)

      (0,0)(-26,-5) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (8.0,0.5)\(17^{\circ}\) (1.3,0.5)\(70^{\circ}\) (5.8,4.4)\(93^{\circ}\)

      \(\)
      The complement of \(70^{\circ}\) is \(90-70=20^{\circ}\).

      The supplement of \(70^{\circ}\) is \(180-70=110^{\circ}\).

    3. Two rural roads intersect. One of the acute angles formed by the roads is \(53^{\circ}\). Find the measures of the other three angles.

      Solution:

      Preamble:

      (0,2)(-18,-4) (0,0)(3,1)6 (0,0)(-3,-1)6 (0,0)(3,-1)6 (0,0)(-3,1)6 (-0.2,0.5)\(C\) (-0.8,-1.0)\(B\) (1.5,-0.3)\(A\) (-3.2,-0.3)\(53^{\circ}\) (3.0,1) (-3.0,1) (3.0,-1) (-3.0,-1) (3.0,1.5)\(1\) (-3.0,1.5)\(2\) (3.0,-2.0)\(3\) (-3.0,-2.0)\(4\)

      \(\)
      Equation:

      The four angles make a complete revolution (\(360^{\circ}\)). \(A\) added to \(B\) measures a half revolution or \(180^{\circ}\).

      Similarly \(B+53=180^{\circ}\). Thus \(B=180-53=127^{\circ}\). \(B\) is the supplement of \(53^{\circ}\).

      \(C+53=180^{\circ}\). Thus \(C=180-53=127^{\circ}\). \(C\) is the supplement of \(53^{\circ}\).

      \(C+A=180^{\circ}\). Thus \(A=180-C=180-127=53^{\circ}\).

    4. A pair of railroad tracks (parallel lines) is intersected by a road (transversal). A gate at \(C\) is to swing between the track and the road. Through how many degrees should the gate swing?

      (0,4)(-25,-6) (0,0)(1,0)20 (0,0)(-1,0)3 (0,6)(1,0)20 (0,6)(-1,0)3 (0,0)(2,1)18 (0,0)(-2,-1)4 (11.5,6)(-2,-1)12 (11.45,6)(-2,-1)12 (11.4,6)(-2,-1)12 (11.35,6)(-2,-1)12 (11.3,6)(-2,-1)12 (11.25,6)(-2,-1)12 (11.2,6)(-2,-1)12 (11.15,6)(-2,-1)12 (0,-1.5)\(A\) (1.9,0.3)\(80^{\circ}\) (8.2,5.0)\(C\) (14.0,6.3)\(B\)

      \(\)
      Solution:

      \(80^{\circ}\) added to \(A=180^{\circ}\). \(A=100\) is the supplement of \(80^{\circ}\) .

      \(80^{\circ}\) and \(C\) are alternate interior angles. Their measures are equal. Thus \(C=80^{\circ}\).

      The gate should swing through \(80^{\circ}\).

    5. A bridge crosses a river at an angle. Angular building \(1\) sits on one side of the bridge. Angular building \(2\) is constructed on the other side of the bridge. The angle for building \(2\) is twice that of building \(1\). The angles left between the buildings and the bridge are shown in the figure on the right. Find the measure of the acute angle between the bridge and the river. \(\)

      (0,7)(-1,-3) (10,6)(1,0)8 (10,6)(-1,0)7 (10,1)(1,0)8 (10,1)(-1,0)7 (11.5,6)(-2,-1)12 (11.5,6.1)(-2,-1)12 (11.5,6.2)(-2,-1)12 (11.5,6.3)(-2,-1)12 (11.5,6)(2,1)7 (11.5,6.1)(2,1)7 (11.5,6.2)(2,1)7 (11.5,6.3)(2,1)7 (11.5,6)(-2,-1)1 (11.5,6.1)(-2,-1)1 (11.5,6.2)(-2,-1)1 (11.5,6.3)(-2,-1)1 (11.5,6.4)(4,1)8 (11.5,6.0)(-1,2)3 (12.4,2.9)river (1.4,3.0)bridge (16.9,6.9)angular building \(1\) (4.0,9.5)angular (4.0,8.5)building \(2\) (8.0,6.2)\(2x^{\circ}\) (14.4,6.2)\(x^{\circ}\) (17.5,8.2)\(18^{\circ}\) (11.0,8.0)\(36^{\circ}\) (19.0,3)not drawn to scale

      \(\)
      Solution:
      The two adjacent angles on the river’s edge form a straight angle. The sum of their measures is \(180^{\circ}\).
      \(\begin{array}{rcl lll} (2x+36)+(x+18)&=&180\\[5pt] 2x+72+x+18&=&180\\[5pt] 3x+90&=&180\\[5pt] 3x+90-90&=&180-90\\[5pt] 3x&=&90\\[15pt] \displaystyle \frac{3}{3}x&=&\displaystyle \frac{90}{3}\\[15pt] x&=&30 \end{array}\)

      (0,3)(-6,4) (10,6)(1,0)10 (10,6)(-1,0)2 (11.5,6)(2,1)7 (14.0,6.3)\(30+18=48^{\circ}\)

      \(\)
      The angle between the bridge and the road is \(48^{\circ}\)

    6. A log is in a horizontal position. It is suspended by two cables joined at a point to form an isosceles triangle with the log as one side of the triangle. Each cable is \(5\) ft less than \(3\) times the length of the log. Find the length of one cable if the perimeter is \(95\) ft.
      Solution:

      Preamble:

      See the picture.

      (0,0)(-20,1) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (4.5,-1.0)\(x\) (-0.6,3.5)\(3x-5\) (8.8,3.5)\(3x-5\)

      \(\)
      Equation:

      \(\begin{array}{rcl lll} \hbox{Perimeter}&=&\hbox{Perimeter}\\[5pt] \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!x\!+\!(3x\!-\!5)\!+\!(3x\!-\!5)&=&95\\[5pt] x+3x-5+3x-5&=&95\\[5pt] 7x-10&=&95\\[5pt] 7x-10+10&=&\!95\!+\!10\\[5pt] 7x&=&105\\[5pt] \displaystyle \frac{7}{7}x&=&\displaystyle \frac{105}{7}\\[15pt] x&=&15\\[5pt] \end{array}\)

      (0,0)(-20,-7) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (4.5,-1.0)\(15\) (11.4,4.5)\(3(15)-5\) (11.4,3.0)\(=45-5\) (11.4,1.5)\(=40\) (1.2,3.5)\(40\)

      \(\)
      Each cable is \(40\) ft long.

    7. A block of wood has a cross section in the form of an equilateral (all three sides are congruent) triangle \(ABC\) as shown on the right. Each side is \(10\) ft long. Trapezoid \(ADEC\) is formed by cutting off the top half of the triangle. What is the length of the top \(DE\) of the trapezoid?

      (0,0)(-24,-6) (0,0)(1,0)10.2 (0,0)(1,1)7 (7,7)(1,-2)3.5 (3.5,3.5)(1,0)5 (-1.0,-1.0)\(A\) (6.4,7.5)\(B\) (9.8,-1.0)\(C\) (3.5,3.5) (8.7,3.5) (2.0,3.0)\(D\) (9.4,3.0)\(E\) (5.4,-1.0)\(10\) (0.6,1.5)\(5\) (10.2,1.5)\(5\) (4.0,5.5)\(5\) (7.9,5.5)\(5\)

      \(\)

      (0,0)(-37,-7.4) (0,0)(1,0)10.2 (3.5,3.5)(-1,-1)3.5 (8.7,3.5)(1,-2)1.7 (3.5,3.5)(1,0)5 (-1.0,-1.0)\(A\) (9.8,-1.0)\(C\) (3.5,3.5) (8.7,3.5) (2.0,3.0)\(D\) (9.4,3.0)\(E\)

      \(\)
      Solution:

      The question boils down to triangle \(BDE\). This triangle is also an equilateral triangle. So side \(DE=5\) ft.

    8. A pile of logs is stacked in the triangular shape shown in the triangle on the right.

      Side \(1\) of a triangle is \(3\) ft more than side \(2\). Side \(3\) is \(3\) ft. less than side \(2\).

      The side of a square is \(15\) ft. The perimeter of the triangle and the square is the same number.

      What are the dimensions of the triangle?

      Solution:

      Preamble:

      (0,0)(-9,1) (0,0)(1,0)11.5 (0,0)(2,3)3.6 (11.5,0)(-3,2)8.0 (5.0,-1.0)\(x\) (-1.9,3.0)\(x-3\) (9.0,3.0)\(x+3\) (25.5,2.0)\(15\) (22,-1.0)\(15\) (20,0)(1,0)5 (20,0)(0,1)5 (25,5)(-1,0)5 (25,5)(0,-1)5 (1,0.5)(1,0)10 (1.5,1.5)(1,0)8 (2.0,2.5)(1,0)6 (2.5,3.5)(1,0)4 (3.0,4.5)(1,0)2 (20.5,0.5)(1,0)5 (20.5,1.5)(1,0)5 (20.5,2.5)(1,0)5 (20.5,3.5)(1,0)5 (20.5,4.5)(1,0)5

      \(\)
      Let \(x\) be side \(2\). Then \(x+3\) and \(x-3\) are the other sides of the triangle

      Perimeter of triangle:
      \((x-3)+x+(x+3)=x-3+x+x+3=3x\)

      Perimeter of square: \(4(15)=60\)

      Equation:

      (0,0)(-23,6) (0,0)(1,0)11.5 (0,0)(2,3)3.6 (11.5,0)(-3,2)8.0 (5.0,-1.0)\(20\) (-1.1,3.0)\(17\) (9.4,3.0)\(23\) (1,0.5)(1,0)10 (1.5,1.5)(1,0)8 (2.0,2.5)(1,0)6 (2.5,3.5)(1,0)4 (3.0,4.5)(1,0)2

      \(\)
      \(\begin{array}{rcl lll} \hbox{Perimeter of triangle}&=&\hbox{Perimeter of square}\\ 3x&=&60\\[10pt] \displaystyle \frac{3}{3}x&=&\displaystyle \frac{60}{3}\\[10pt] x&=&20 \end{array}\)

      The dimensions of the triangle are

      \(20-3=17\), \(20\) and \(20+3=23\) ft.

    9. Two searchlights are programmed to illuminate the sky at angles configured in the picture on the right. Find the three angles.

      (0,5)(-20,4) (0,0)(1,0)5.0 (0,0)(-1,0)5 (0,0)(-1,2)3.0 (0,0)(1,1)5.5 (3.5,2.5)\(3x+17\) (-6.6,2.5)\(5x-6\) (-0.8,2.5)\(x+7\) (3.5,7.0)searchlight 1 (-8.5,7.0)searchlight 2

      Solution

      Preamble:

      The sum of the three angles is a straight angle.

      Equation:

      \(\begin{array}{rcl lll} \hbox{The sum of the three angles}&=&180^{\circ}\\[5pt] \!(5x\!-\!6)\!+\!(x\!+\!7)\!+\!(3x\!+\!17)&=&180\\[5pt] 5x-6+x+7+3x+17&=&180\\[5pt] 9x-6+7+17&=&180\\[5pt] 9x+18&=&180\\[5pt] 9x+18-18&=&180-18\\[5pt] 9x&=&162\\[15pt] \displaystyle \frac{9}{9}x&=&\displaystyle \frac{162}{9}\\[15pt] x&=&18\\ \end{array}\)

      \(5x-6=5(18)-6=90-6=84\),

      \(x+7=18+7=25\),

      \(3x+17=3(18)+17=54+17=71\).

      (0,6)(-10,-1) (0,0)(1,0)5.0 (0,0)(-1,0)5 (0,0)(-1,2)2.0 (0,0)(1,1)3.5 (2.5,1.0)\(71^{\circ}\) (-3.1,1.0)\(84^{\circ}\) (-0.4,1.0)\(25^{\circ}\)

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