Math 3 Week 6: Solving Triangles Using Trigonometry

Math 3 Week 6: Solving Triangles Using Trigonometry

11th Grade

12 Qs

quiz-placeholder

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Math 3 Week 6: Solving Triangles Using Trigonometry

Math 3 Week 6: Solving Triangles Using Trigonometry

Assessment

Quiz

Mathematics

11th Grade

Medium

CCSS
HSG.SRT.D.11, HSG.SRT.C.8, HSG.SRT.C.6

+1

Standards-aligned

Created by

Kylie Campbell

Used 7+ times

FREE Resource

12 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

What would be the BEST way to solve for a?

Pythagorean Theorem

Law of Sines

Law of Cosines

cos a=912\cos\ a=\frac{9}{12}

Tags

CCSS.HSG.SRT.C.6

CCSS.HSG.SRT.C.8

CCSS.HSG.SRT.D.11

2.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

What would be the BEST way to solve for c?

Law of Sines

Law of Cosines

 sin25°=bc\sin25\degree=\frac{b}{c}  

 cos25°=8c\cos25\degree=\frac{8}{c} 

Tags

CCSS.HSG.SRT.C.8

CCSS.HSG.SRT.D.11

3.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

What would be the BEST way to solve for c?

Pythagorean Theorem

Law of Sines

Law of Cosines

 sin37°=c11\sin37\degree=\frac{c}{11} 

Tags

CCSS.HSG.SRT.C.8

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

4.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

Media Image

What would be the BEST way to solve for p?

Pythagorean Theorem

Law of Sines

Law of Cosines

 sin32°=p21\sin32\degree=\frac{p}{21}  

Tags

CCSS.HSG.SRT.C.6

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

5.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Media Image

Which two ratios would you need to use the Law of Sines to find b? (Select 2)

 sin50°b\frac{\sin50\degree}{b}  

 sinAa\frac{\sin A}{a}  

 sin110°12\frac{\sin110\degree}{12}  

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

6.

MULTIPLE SELECT QUESTION

2 mins • 1 pt

Media Image

Which two ratios would you need to use the Law of Sines to find a? (Select 2)

 sin50°b\frac{\sin50\degree}{b}  

 sinAa\frac{\sin A}{a}  

 sin110°12\frac{\sin110\degree}{12}  

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

Which of the following is the correct set up of the Law of Sines formula for this triangle?

 sin20°a=sin50°b=sin110°12\frac{\sin20\degree}{a}=\frac{\sin50\degree}{b}=\frac{\sin110\degree}{12}  

 sinAa=sinb50=sin110°12\frac{\sin A}{a}=\frac{\sin b}{50}=\frac{\sin110\degree}{12}  

 sin20°A=sin50°b=sin12110°\frac{\sin20\degree}{A}=\frac{\sin50\degree}{b}=\frac{\sin12}{110\degree}  

 sinAa=sin110°b=sin50°12\frac{\sin A}{a}=\frac{\sin110\degree}{b}=\frac{\sin50\degree}{12}  

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

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