Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Line (geometry) wikipedia , lookup
Technical drawing wikipedia , lookup
Rotation formalisms in three dimensions wikipedia , lookup
Perceived visual angle wikipedia , lookup
Pythagorean theorem wikipedia , lookup
Multilateration wikipedia , lookup
History of trigonometry wikipedia , lookup
Integer triangle wikipedia , lookup
Rational trigonometry wikipedia , lookup
Compass-and-straightedge construction wikipedia , lookup
Trigonometric functions wikipedia , lookup
b) Statement S(n) = 180°(n – 2) S(8) = 180°(8 – 2) S(8) = 180°(6) S(8) = 1080° 1080º = 135° 8 x + 135° = 180° x = 45° Justification A regular octagon has eight sides, so n is 8. Measure of one interior angle. Let x be the measure of an exterior angle. Supplementary angles Cha pt e r R e vie w, pa ge 1 0 6 c) Statement AHG = 135° and FGH = 135° HAF = 45° and GFA = 45° HAB = 135° HAB = HAF + FAB FAB = HAB – HAF FAB = 135° – 45° FAB = 90° Similarly, ABE = 90° AF || BE Let S represent the sum of the angles indicated. S = 2(a + b + c + d + e + f) S = 2(360°) S = 720° The sum of the indicated angles is 720°. Alternative solution: The angles are external angles of the central hexagon. Since there are two external angles at each vertex, the sum of the angles is double the sum of the external angles of a polygon, or 2(360°) = 720°. Justification Regular octagon Exterior angle of regular octagon Regular octagon Addition and substitution Solve for FAB. Symmetry Supplementary angles 6. e.g., Label the indicated angles a, b, c, d, e, and f. As a result, the measures of the supplementary angles are as shown. The interior angles in a hexagon sum to 720°. Therefore, 720° = (180° – a) + (180° – b) + (180° – c) + (180° – d) + (180° – e) + (180° – f) 720° = 1080° – (a + b + c + d + e + f) 360° = a + b + c + d + e + f By opposite angles, each angle that is opposite to a, b, c, d, e, and f is equal to that angle, as shown. 1. e.g., The side bars coming up to the handle are parallel and the handle is a transversal. 2. a) Corresponding pairs of angles: a, e; b, g; c, f; d, h b) e.g., No. The lines are not parallel, so corresponding pairs cannot be equal. c) I can see eight pairs of supplementary angles. a, b; a, c; c, d; b, d; e, f; e, g; f, h; g, h d) The other pairs of equal angles are vertically opposite angles: a, d; b, c; e, h; f, g 3. Statement a = 35° b + a = 180° b = 145° 4. Let b be the 180° angle, a be the supplementary angle to b, and c be the 40° angle. Statement a+b a c a AB = 180° = 40° = 40° =c || CD Justification Supplementary angles Substitution Given Transitive principle Corresponding angles are equal. 5. a) Statement b = 76° c = 76° a + b = 180° a = 104° Foundations of Mathematics 11 Solutions Manual Justification Corresponding angles Supplementary angles Justification Corresponding angles Vertically opposite angles Supplementary angles 2-15 b) 9. Statement 2a + 3a 5a a 3a 3a 2a 2a b b c + 2a c + 72° c = 180° = 180° = 36° = 3(36°) = 108° = 2(36°) = 72° = 3a = 108° = 180° = 180° = 108° Justification Interior angles on the same side of a transversal Substitution Vertically opposite angles Supplementary angles Substitution b) Measure ABF and BFH. Measure DBA and BFE. Both pairs should be equal. So then the lines are parallel. 7. ST = TR Justification Given angles Alternate interior angles Transitive property Base angles of TRS are equal, so TRS is isosceles. Property of isosceles triangle 8. a) Statement x = 40° z = 45° x + y + z = 180° 40° + y + 45° = 180° y = 95° OPL NO Substitution 6. e.g., a) Statement QRS = TRS QRS = RST TRS = RST Statement LO = LP Justification Alternate interior angles Alternate interior angles Supplementary angles Substitution OQN Justification Given LOP is isosceles. Property of isosceles = POL triangle = NQ Given NOQ is isosceles. Property of isosceles = NOQ triangle Sum of interior = PLO + POL + OPL angles in = 180° – (POL + OPL) triangle = 180° – 2POL 180° PLO PLO Sum of interior 180° = QNO + NOQ + OQN angles in QNO = 180° – (NOQ + OQN) triangle QNO = 180° – 2NOQ PLO and QNO are PLO + QNO = 180° – 90° the two acute angles PLO + QNO = 90° in right triangle LMN. Sum of 180° – 2POL + 180° – 2NOQ = 90° interior 360° – 2(POL + NOQ) = 90° –2(POL + NOQ) = –270° angles of triangle is POL + NOQ = 135° 180°. Substitution POQ + POL + NOQ = 180° Supplementary POQ + 135° = 180° angles Substitution POQ = 45° 10. a) Statement S(n) = 180°(n – 2) S(15) = 180°(15 – 2) S(15) = 180°(13) S(15) = 2340° Justification For a 15-sided polygon, n is 15. b) The sum of the exterior angles of any polygon is 360°. The measure of one exterior angle is 360° divided by the number of sides. 360° = 24° 15 The measure of an exterior angle of a 15-sided polygon is 24°. b) Statement y + 68° = 180° y = 112° x = 68° z + 72° + 68° = 180° z = 40° 2-16 Justification Supplementary angles Corresponding angles Sum of interior angles in triangle Chapter 2: Properties of Angles and Triangles 11. Statement S(n) = 180°(n – 2) S(5) = 180°(5 – 2) S(5) = 180°(3) S(5) = 540° 540º = 108° 5 AB = BC BAC Justification A regular pentagon has five sides, so n is 5. Measure of one interior angle Given ABC is isosceles. Property of isosceles = BCA triangle Sum of = BAC + BCA + interior angles in = BAC + BCA + 108° triangle = BAC + BCA Substitution = 2BAC = BAC Property of = BCA + ACD equality = BCD – BCA 180° ABC 180° 72° 72° 36° BCD ACD ACD = 108° – 36° ACD = 72° 180° = ACD + CDE Sum of interior angles on same side of transversal AC || ED Cha pt e r Ta s k, pa ge 10 7 e.g., A. I decided to create a logo for my local animal shelter, to raise awareness of homeless animals. I want to convey the message that dogs are good for you and give you love. If you’re looking for love, come to the animal shelter! B. I chose a stylized heart, using parallel lines as thin stripes to form the heart, because I think this design is more interesting than a solid heart. I wanted to include a doghouse as well, to show that animals live here. Again, I wanted my design to be a bit different, to catch the eye, so I chose a pentagon. I lined up the heart and the pentagon so that one edge of each is parallel to the other. C. I started my logo by drawing a horizontal line. Then I used a protractor to draw a line at an angle of 26° from the horizontal. I drew a vertical line at the point of intersection of these two lines. The angle between the vertical line and the line drawn at a 26° angle should have a measure of 64°. I measured it to be sure. Next, I drew the H. I drew a line above the H, parallel to the line at the base of the H, by measuring a 64° angle to the vertical line (corresponding angles). I completed the words and then drew another vertical line at the end of the a. Foundations of Mathematics 11 Solutions Manual Next, I printed out an outline of a heart. I placed the heart on top of a piece of graph paper and used the graph paper lines as a guide to draw the parallel lines for the stripes. When I finished, I checked my work. I used a straightedge as a transversal, and I made sure that each line was lined up with the straightedge. Then I rotated the heart into the position I wanted, under the words. Finally, I used dynamic geometry software to draw a regular pentagon. I determined that the sum of the measures of the interior angles would be 540°. S(n) = 180°(n – 2) S(5) = 180°(5 – 2) S(5) = 540° This meant that the measure of each interior angle 540° would be = 108° . 5 I adjusted the vertices until all my angles were 108°. I knew that the sides would all have the same measure. I printed the pentagon. I added extra lines, as shown, for the roof. D. To make the final form of my logo, I erased the lines and measures in my sketch. I drew a dog, sized it to fit inside the doghouse, and glued it onto the house. Then I scanned the completed logo and touched it up. 2-17