Unit 3 Assessment The Trigonometry of General Triangles

Which method would be used to solve this triangle?

• Law of Sines

• Law of Cosines

Which method would be used to solve this triangle?

• Law of Sines

• Law of Cosines

Solve the Triangle(s):  $a=14$ ,  $b=6$ , $A=27\degree$  

• no triangle possible

• B = 11.219^o, C = 141.781^o, c = 19.078

• B = 38.219^o, C = 114.781^o, c = 27.998

• B = 11.219^o, C = 141.781^o, c = 19.078

  OR

  B = 38.219^o, C = 114.781^o, c = 27.998

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

Solve the Triangle(s): A = 26^o, a = 5, b = 9

• no triangle possible

• B = 14^o, C = 21^o, c = 8.7

• B = 52^o, C = 102^o, c = 11.2

• B = 52^o , C = 102^o, c = 11.2

  OR

  B = 128^o, C = 26^o, c = 5.0

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

Identify the Law of Sines

• Sin²x + Cos²x =1
• Sin(A)/a = Sin(B)/b = Sin(C)/c
• Sin(A)Sin(B)Sin(C) = 1
• Soh-Cah-Toa

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

Which of the following formulas shows the Law of Cosines?

• c² = a² + b² - 4ac + cosA
• c² = a² - b² - 2abcosC
• c² = a² + b² - 2abcosC
• (cos A)/a = (cos B)/b

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

Solve the triangle

• b=2.95 , A=28.7 , C=106.3
• b= 2.95, A= 30.6, C=108.43
• b=0.67 , A=28.7 , C=105.78
• b=4.95 , A= 21, C=106

Tags

CCSS.HSG.SRT.C.8

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

Solve the Triangle

a=38, b= 31, c= 35

Find <A, <B and <C

• A= 43°, B= 47°, C= 90°

• A= 70°, B= 50.1°, C= 59.9°

• A= 50.1°, B= 70°, C= 59.9°

• A= 60°, B= 70°, C= 50°

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

What are the possible criteria for Law of Sines?

• SSA, ASA, AAS

• AAS, SSS, ASA

• SAS, SSS, ASA

• SAS, AAS, ASA

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

What are the possible criteria for the Law of Cosines?

• SSA, SSS

• SSA, ASA

• SSS, SAS

• AAS, ASA

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11