Math 3 Week 6: Solving Triangles Using Trigonometry

What would be the BEST way to solve for a?

• Pythagorean Theorem

• Law of Sines

• Law of Cosines

• $\cos\ a=\frac{9}{12}$

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

What would be the BEST way to solve for c?

• Law of Sines

• Law of Cosines

•  $\sin25\degree=\frac{b}{c}$  

•  $\cos25\degree=\frac{8}{c}$ 

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

What would be the BEST way to solve for c?

• Pythagorean Theorem

• Law of Sines

• Law of Cosines

•  $\sin37\degree=\frac{c}{11}$ 

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

What would be the BEST way to solve for p?

• Pythagorean Theorem

• Law of Sines

• Law of Cosines

•  $\sin32\degree=\frac{p}{21}$  

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

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

•  $\frac{\sin50\degree}{b}$  

•  $\frac{\sin A}{a}$  

•  $\frac{\sin110\degree}{12}$  

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

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

•  $\frac{\sin50\degree}{b}$  

•  $\frac{\sin A}{a}$  

•  $\frac{\sin110\degree}{12}$  

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

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

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

•  $\frac{\sin A}{a}=\frac{\sin b}{50}=\frac{\sin110\degree}{12}$  

•  $\frac{\sin20\degree}{A}=\frac{\sin50\degree}{b}=\frac{\sin12}{110\degree}$  

•  $\frac{\sin A}{a}=\frac{\sin110\degree}{b}=\frac{\sin50\degree}{12}$  

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

Find x.

Round your answer to the nearest degree.

• $53\degree$

• $37\degree$

• $49\degree$

• $1.33\degree$

Tags

CCSS.HSG.SRT.C.8

Find x.
Round your answer to the nearest tenths place.

•  $22.6$ 

•  $6.4$ 

•  $14.2$ 

•  $12$ 

Tags

CCSS.HSG.SRT.C.8

Which of the following is the correct way to set up the Law of Cosines to find the missing side length?

• $x^2=200^2+300^2-\left(200\right)\left(300\right)\left(\cos40\degree\right)$

• $x^2=200^2+300^2+\left(200\right)\left(300\right)\left(\cos40\degree\right)$

• $x^2=200^2-300^2-\left(200\right)\left(300\right)\left(\cos40\degree\right)$

• $40^2=200^2+300^2-\left(200\right)\left(300\right)\left(\cos x\degree\right)$

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11