Chapter 7 - Geometry and Trigonometry Study Guide

Chapter 7: Topics Overview

7.1 - Dimensional Analysis

Dimensional analysis involves converting between different units of measurement. It is a critical skill in many scientific fields, particularly physics and chemistry, where measurements often need to be converted from one system of units to another. The process relies on the principle of using conversion factors, which are ratios of equivalent measurements.

Example:

To convert meters to centimeters, one would use the conversion factor of 1extm=100extcm1 ext{ m} = 100 ext{ cm}. Therefore, if you have a measurement of 2 meters and you want to convert it to centimeters:
2extmimesrac100extcm1extm=200extcm2 ext{ m} imes rac{100 ext{ cm}}{1 ext{ m}} = 200 ext{ cm}.

7.2 - Angles

7.2.1 - Complementary & Supplementary Angles

Complementary angles are two angles whose measures add up to 90 degrees. In contrast, supplementary angles are two angles whose measures add up to 180 degrees.

7.2.2 - Vertical Angles

Vertical angles are the opposite angles formed when two lines intersect. These angles are always congruent.

7.2.3 - Alternate Interior Angles, Alternate Exterior Angles, & Corresponding Angles

  • Alternate Interior Angles: Angles that are located between two parallel lines and on opposite sides of a transversal.

  • Alternate Exterior Angles: Angles that lie outside the parallel lines and on opposite sides of a transversal.

  • Corresponding Angles: Angles that are in the same position on different parallel lines cut by a transversal.

7.2.4 - Triangle Angle Sum

The sum of the interior angles of a triangle is always equal to 180 degrees. This foundational property can be used to find missing angle measures within triangles by applying the equation:
extAngle1+extAngle2+extAngle3=180extoext{Angle}_1 + ext{Angle}_2 + ext{Angle}_3 = 180^ ext{o}.

7.3 - Finding Perimeter and Circumference

7.3.1 - Perimeter

To find the perimeter of various shapes:

  • Rectangle: P=2imes(Length+Width)P = 2 imes (Length + Width).

  • Triangle: P=a+b+cP = a + b + c where a, b, and c are the lengths of the sides.

7.3.2 - Circumference

Circumference of a circle can be determined with the formula:
C=extDiameterimesextπC = ext{Diameter} imes ext{π}, where extπext(Pi)extisapproximately3.14ext{π} ext{ (Pi)} ext{ is approximately } 3.14. Alternatively, if the radius (r) is known, it can also be calculated using:
C=2imesextπimesrC = 2 imes ext{π} imes r.

7.4 - Similar Triangles

7.4.1 - Solving for a Missing Side

In similar triangles, corresponding sides are proportional. If triangle ABC is similar to triangle DEF, then:
racABDE=racACDF=racBCEFrac{AB}{DE} = rac{AC}{DF} = rac{BC}{EF}.
Using cross multiplication can help find any missing length in similar triangles.

7.4.2 - Pythagorean Theorem

For right triangles only, the Pythagorean theorem states that:
a2+b2=c2a^2 + b^2 = c^2, where c represents the length of the hypotenuse, and a and b are the lengths of the other two sides. This theorem is crucial for determining missing length values in right triangles.

7.5 - Finding Volume

7.5.1 - Volume of Various Shapes

  • Rectangular Prism: V=LengthimesWidthimesHeightV = Length imes Width imes Height.

  • Cylinder: V=extπimesr2imeshV = ext{π} imes r^2 imes h where r is the radius and h is the height.

7.6 - Finding Surface Area and Trigonometric Ratios

7.6.1 - Surface Area

Formulas for surface area depend on the shape:

  • Cube: SA=6s2SA = 6s^2 where s is the length of a side.

  • Rectangular Prism: SA=2lw+2lh+2whSA = 2lw + 2lh + 2wh.

7.6.2 - Finding Trigonometric Ratios

In right triangles, the common trigonometric ratios are defined as:

  • Sine: extsin(heta)=racextoppositeexthypotenuseext{sin}( heta) = rac{ ext{opposite}}{ ext{hypotenuse}}.

  • Cosine: extcos(heta)=racextadjacentexthypotenuseext{cos}( heta) = rac{ ext{adjacent}}{ ext{hypotenuse}}.

  • Tangent: exttan(heta)=racextoppositeextadjacentext{tan}( heta) = rac{ ext{opposite}}{ ext{adjacent}}.

7.6.3 - Solving for Side or Angle

Using trigonometric ratios, one can either solve for an unknown side or angle in a right triangle. Given an angle, if one side is known, use appropriate trigonometric functions to find missing lengths.

If an angle is known, to solve for a side, apply:
extopposite=exthypotenuseimesextsin(heta)ext{opposite} = ext{hypotenuse} imes ext{sin}( heta), etc.

Practical Applications

Example Problems

  1. Find the length of hedge surrounding a rectangular playground of length 160 ft and width 120 ft:
    P=2imes(160+120)=560extft.P = 2 imes (160 + 120) = 560 ext{ ft}.

  2. Find the circumference of a circle with a diameter of 4.5 m:
    C=4.5imes3.14ext(approx)=14.1m.C = 4.5 imes 3.14 ext{ (approx)} = 14.1 m.

  3. Find the land area of a trapezoid-shaped township with bases measuring 9 km and 23 km and a height of 10 km:
    A=rac12imes(b1+b2)imesh=rac12imes(9+23)imes10=160extkm2.A = rac{1}{2} imes (b_1 + b_2) imes h = rac{1}{2} imes (9 + 23) imes 10 = 160 ext{ km²}.

  4. Cost to carpet a rectangular floor of dimensions 12 feet by 15 feet at $18.50 per square yard:
    extArea=12imes15=180extft2<br>ightarrow20extyd2<br>ightarrowextTotalCost=20imes18.50=370.00ext{Area} = 12 imes 15 = 180 ext{ ft²} <br>ightarrow 20 ext{ yd²} <br>ightarrow ext{Total Cost} = 20 imes 18.50 = 370.00.

  5. Find the hypotenuse length in a right triangle whose legs are 7 feet and 24 feet:
    c=ext√(72+242)=ext√(49+576)=25extft.c = ext{√}(7^2 + 24^2) = ext{√}(49 + 576) = 25 ext{ ft}.

  6. Volume of a figure with known dimensions (provide specific dimensions for calculation):
    Volume will depend on the shape in question.

  7. Find the angle of elevation of the Sun given a building height of 21 meters and a shadow length of 25 meters:

  • Use tan function, exttan(heta)=rac2125ext{tan}( heta) = rac{21}{25}, then calculate the angle using inverse tangent.