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EXAMPLE Solve for the missing side of the triangle shown in the figure below. Solution: Find a. a² = b²+c² - 2bc cos A A 100.67 c = 10 in b=7.98 in a=√(7.98)2 + (10)2-2(7.98) (10) cos 100.67° a = √63.6804+ 100-159.6 cos 100.67° a = 163.6804- 159.6(-0.1851521)=√163.6804 + 29.55027 = a≈ 13.9 in Solve for R € to aglenomia a w sans tooll nabod were At B C a √193.23067

EXAMPLE Solve for the missing side of the triangle shown in the figure below. Solution: Find a. a² = b²+c² - 2bc cos A A 100.67 c = 10 in b=7.98 in a=√(7.98)2 + (10)2-2(7.98) (10) cos 100.67° a = √63.6804+ 100-159.6 cos 100.67° a = 163.6804- 159.6(-0.1851521)=√163.6804 + 29.55027 = a≈ 13.9 in Solve for R € to aglenomia a w sans tooll nabod were At B C a √193.23067

Elementary Geometry For College Students, 7e
Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 1CT
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LEVEL LAS TITLE LEARNING COMPETENCY DISTRICT Basic Literacy Lower Elementary Advanced Elementary Junior High School Senior High School Law of Cosines LAS NO. 19 SCORE Illustrate the Law of Cosines. 1. CONCEPT NOTES • The Law of Cosines is an extension of Pythagoras' theorem on non-right angle triangle. • Also known as the cosine rule, it says that if a triangle has sides of lengths a, b, and c, and with opposite angles A, B, and C, respectively, then a² = b² + c² 2bc cos A 1 ⚫ The Law of Cosines is used in solving triangles when given a. Two sides and the included angle (SAS) b. Three sides (SSS) EXAMPLE Solve for the missing side of the triangle shown in the figure below. Solution: Find a. a² = b² + c² - 2bc cos A a=√(7.98)2 + (10) 2 - 2(7.98) (10) cos 100.67° a = √63.6804+ 100-159.6 cos 100.67° B A 100.67 c = 10 in b=7.98 in a a = √163.6804-159.6(-0.1851521) = √163.6804 + 29.55027 = √193.23067 C a = 13.9 in Solve for B. D a sin B sin A b sin A sin B a sin B = 7.98(sin 100.67°) 13.9 sin B 0.564 Bsin 0.564 B≈ 34.34° II. EXERCISES Solve the following: (10 pts) 2 130 Solve for C. C=180°- (100.67° +34.34°) agld we wer C≈44.99° Sambrue final a anil 098A slur a C Solve for the missing side of the triangle shown in the figure on the right. Given, b = 6.83 in, c = 9 in and A = 102.38° b = 6.83 in c = 9 in 102.38 Note This is just a template which you can use as reference when you craft your CVIF-DLP Learning Activity Sheets A vivo T1x B
Transcribed Image Text:LEVEL LAS TITLE LEARNING COMPETENCY DISTRICT Basic Literacy Lower Elementary Advanced Elementary Junior High School Senior High School Law of Cosines LAS NO. 19 SCORE Illustrate the Law of Cosines. 1. CONCEPT NOTES • The Law of Cosines is an extension of Pythagoras' theorem on non-right angle triangle. • Also known as the cosine rule, it says that if a triangle has sides of lengths a, b, and c, and with opposite angles A, B, and C, respectively, then a² = b² + c² 2bc cos A 1 ⚫ The Law of Cosines is used in solving triangles when given a. Two sides and the included angle (SAS) b. Three sides (SSS) EXAMPLE Solve for the missing side of the triangle shown in the figure below. Solution: Find a. a² = b² + c² - 2bc cos A a=√(7.98)2 + (10) 2 - 2(7.98) (10) cos 100.67° a = √63.6804+ 100-159.6 cos 100.67° B A 100.67 c = 10 in b=7.98 in a a = √163.6804-159.6(-0.1851521) = √163.6804 + 29.55027 = √193.23067 C a = 13.9 in Solve for B. D a sin B sin A b sin A sin B a sin B = 7.98(sin 100.67°) 13.9 sin B 0.564 Bsin 0.564 B≈ 34.34° II. EXERCISES Solve the following: (10 pts) 2 130 Solve for C. C=180°- (100.67° +34.34°) agld we wer C≈44.99° Sambrue final a anil 098A slur a C Solve for the missing side of the triangle shown in the figure on the right. Given, b = 6.83 in, c = 9 in and A = 102.38° b = 6.83 in c = 9 in 102.38 Note This is just a template which you can use as reference when you craft your CVIF-DLP Learning Activity Sheets A vivo T1x B
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