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## Geometry

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**Geometry:Part IIIByDick Gill, Julia Arnold and Marcia**Tharpfor Elementary Algebra Math 03 online**Table of Contents**Converting Degrees to Degrees and Minutes Converting Degrees and Minutes to decimal degrees Vertical Angles Straight Angles Parallel Lines Problems involving the above and similar triangles.**Fractions of Angles**One degree is an awfully small angle, but when we need to talk about a fraction of a degree, we can do so with decimals or with minutes and seconds. One degree can be divided into 60 minutes and one minute can be divided into 60 seconds. The notation for minutes and seconds will look like this: 1o = 60’ 1’ = 60” 25.5o = 25o 30’ Which brings up the question, how do you convert from decimal fractions to fractions with minutes and seconds?**Converting from Decimals to Minutes**To convert 57.4o degrees to degrees and minutes we can write the degrees as 570 + .4o . Next we find 0.4 of 60’minutes by multiplying 0.4(60’) = 24’minutes . Therefore, 57.4o = 57o 24’. Try a few of these on your own. Convert each of the following to degrees and minutes: 123.8o = 123o + 0.8(60’) = 123o + 48’ = 123o 48’ = 12o + 0.3(60’) = 12o + 18’ = 12o 18’ 12.3o = 82o + 0.33(60’) = 82o + 19.8’ which rounds to 82o 20’ Though we won’t focus on it in this course we could convert 19.8’ to 19’ 48”. 82.33o**Converting from Degrees and Minutes to Decimal Degrees**To convert the minutes to a decimal it is helpful to remember that minutes represent a fraction of a degree. We need to get that fraction into decimal form. For example in120o 45’ the 45’ are the fraction of a degree: (45/60)o = .75o so 120o 45’ = 120.75o. See what you can do with the following conversions. Round to the nearest one hundredth of a degree. 12o 52’ = 12o + (52/60)o = 12o + .866o rounding to 12.87o 135o 45’ = 135o + (45/60)o = 135o + .75o = 135.75o = 60o + (25/60)o = 60o + .416o rounding to 60.42o 60o 25’**Practice Problems**1. Convert 1680 15’ to degrees. 2. Convert 145.880 to degrees and minutes. Work out the answers then click to the next slide to check.**Answers**1. Convert 1680 15’ to degrees. 1680 +15/60=168 + .25 = 168.250 2. Convert 145.880 to degrees and minutes. 1450 + .88*60= 1450 52.8’ Table of Contents**Vertical Angles**When two lines intersect they form vertical angles. Angles 3 and 4 form a pair of vertical angles 3 Angles 1 and 2 form a pair of vertical angles. 2 1 4**Straight Angles**A straight angle is an angle formed by a straight line and has measure 1800. 1800**Let’s explore**Suppose we have two intersecting lines with angle measures as shown. 60 120 What do these two angles add up to? Do you see the straight angle?**Let’s explore**What should the measure of the ? angle be? 60 120 ? 120 What measure is left for the remaining angle?**Let’s explore**Can you come up with a property about vertical angles after looking at this example? 60 120 120 60 The property is Vertical angles are always equal to each other.**Look at the following picture and give**the measure of all unknown angles. 100 80 80 100 Table of Contents**Parallel lines**Two lines are parallel if they do not intersect. A line which crosses the parallel lines is called a transversal.**Parallel lines**Look at all the angles formed. 1 2 4 3 5 6 8 7**Let’s explore**If I give the measure of angle 1 as 1200, how many other angles can I find? 20 1 2 4 3 5 6 8 7 See how many angles you can find then click to the next slide.**Names of equal angles in picture**120 60 120 60 60 120 60 120 The red angles are called alternate interior angles. What other pair of angles are also alternate interior angles?**Names of equal angles in picture**The red 600 angles are also alternate interior angles. 120 60 120 60 60 120 60 120**Names of equal angles in picture**The red angles pictured are called Alternate exterior angles. 120 60 120 60 60 120 60 120 Can you find another pair of alternate exterior angles?**Names of equal angles in picture**The red angles pictured are called Alternate exterior angles. 120 60 120 60 60 120 60 120 Do you see that the interior angles are between the parallel lines and the exterior angles are on the outside of the parallel lines?**Names of equal angles in picture**The red angles pictured are called Corresponding angles. 120 60 120 60 60 120 60 120 There are 3 more pairs of corresponding angles. Can you find them?**Names of equal angles in picture**The corresponding angles are pictured in matching colors. 120 60 60 120 60 120 60 120**Properties**If two lines are parallel and cut by a transversal then the alternate interior angles are equal, the alternate exterior angles are equal, and the corresponding angles are equal. Note: Corresponding angles are always on the same side of the transversal.**Can you fill in the missing**angles in the following picture and state the reason why? 2 1 3 4 100o 5 7 6**Ang 7= 100 because of vertical angles.**Ang 5 = 80 because 7 & 5 form a straight angle. Ang 6 is 80 because 5 and 6 are vertical and therefore equal. Ang 4 = 100 because alternate interior angles are equal Ang 2 = 100 because of corresponding angles or because of vertical angles. Angle 3 = 80 because of straight angles or because 3 & 5 are alt. Int. angles Ang. 1 = 80 because 3 & 1 are vertical or 1&5 are corresponding or 1 and 6 are alt ext angles. 2 1 3 4 100o 5 7 6 Table of Contents**Problems involving similar triangles,**parallel lines, vertical angles, and equal angles.**C**In this triangle DB || EA This makes CA and CE transversals. What angles are equal? B D CBD = CAE CDB = CEA A E These are two pairs of corresponding angles. Note: The similar tick marks indicate equal angles.**C**What triangles are similar? BCD ___ ACE Which is correct? B D ABC ECA EBA AEC ACE A E**C**Answer the following: AE = 10 DB = 6 1. If DC=7 find CE B D A E When you’ve worked it out click here to check.**C**Answer the following: AE = 10 DB = 6 2. If CA=27 find CB B D A E When you’ve worked it out click here to check**C**Answer the following: AE = 10 DB = 6 3. If CB=10 find BA B D A E When you’ve worked it out click here to check**C**Answer the following: AE = 10 DB = 6 4. If CE=24 find DE B D A E When you’ve worked it out click here to check**Name the equal angles in this figure.**L K = N (Both are right angles) LMK = OMN (Vertical angles) L = O N M K How do you write the similarity of the triangles? NOM KLM Think before you click. O**Click once for the problem.**What are the corresponding sides? KM MN 5. If KM = 6 MN = 9 MO = 12 Find LM LM MO L x M 9 N 6 K 12 When you’ve worked it out click here to check O**6.**A = DBC AC = 12 BC = 8 BD = 5 FIND AB A 12 D 5 8 C B When you’ve worked it out click here to check**B**BD = 5 BE = 8 BA = 10 FIND BC 7. E D A C When you’ve worked it out click here to check**C**Answer the following: AE = 10 DB = 6 1. If DC=7 find CE B D 6CE=70 CE = 35/3 A E Back to problem2**C**Answer the following: AE = 10 DB = 6 2. If CA=27 find CB B D 10CB=162 CB=162/10 CB = 81/5 A E Back to problem3**C**Answer the following: AE = 10 DB = 6 3. If CB=10 find BA B D Back to problem4 We could not use BA because BA is not a side. A E 6CA=100 CA = 100/6=50/3 Now CA - CB = BA or 50/3-10 = (50-30)/3=20/3 BA = 20/3**C**Answer the following: AE = 10 DB = 6 4. If CE=24 find DE Back to problem5 B D A E 10DC=144 DC = 144/10=72/5 Now CE - DC = DE or 24 -72/5= (120-72)/5=48/5 DE = 48/5**5. If KM = 6**MN = 9 MO = 12 Find LM L x N M 9 6 K 12 Back to problem6 O**6.**A = DBC AC = 12 BC = 8 BD = 5 FIND AB A First we must find the similar triangles. 12 What angle is in two triangles? D 5 8 C B**A = DBC**AC = 12 BC = 8 BD = 5 FIND AB A First we must find the similar triangles. 12 What angle is in two triangles? D 5 c 8 C B**B**Imagine separating the two triangles. ABC is flipped up, while BDC is rotated so DC is horizontal. C 8 x 12 12 D C A 5 5 8 A B D ABC BDC 8 C B**ABC BDC**A 12 x D 5 Back to problem7 8 C B**First, what**two triangles are similar? B BD = 5 BE = 8 BA = 10 FIND BC 7. DBE ABC What angle is in both triangles? E D A C**DBE**ABC B BD = 5 BE = 8 BA = 10 FIND BC 7. 8 5 E x 10 D A C Table of Contents Return to Problems End show**You are now ready for the last**geometry topic: Area and Volume Go to Geometry: Part IV