HW1 Pythagorean thm, Special Right Triangles, & Trig Functions

Find the missing length. Give your answer in simplest radical form.

• x = 17

• x = 24

• x = 2√74

• x = 20

Tags

CCSS.8.G.B.7

Find the missing length. Give your answer in simplest radical form.

• x = 12

• x = 5

• x = 10

• x = 2√17

Tags

CCSS.8.G.B.7

Find the missing length. Give your answer in simplest radical form.

• x = 2√10, y = 4√5

• x = 4√10, y = 2√5

• x = 2√5, y = 4√10

• x = 5√2, y = 3√6

Tags

CCSS.HSG.SRT.C.8

Find the missing length. Give your answer in simplest radical form.

• x = 21

  y = 7√2

• x = 3√7

  y = 19

• x = 19

  y = 3√7

• x = 7√3

  y = 21

Tags

CCSS.HSG.SRT.C.8

Find the missing length. Give your answer in simplest radical form.

• x = 10√5

  y = 20√5

• x = 10√3

  y = 20√3

• x = 10√6

  y = 20√6

• x = 10√2

  y = 20√2

Tags

CCSS.HSG.SRT.C.8

Find the missing length. Give your answer in simplest radical form.

• x = 14√5

  y = 14√5

• x = 14√3

  y = 14√3

• x = 14√2

  y = 14√2

• x = 14

  y = 14

Tags

CCSS.HSG.SRT.C.8

Find the missing length. Give your answer in simplest radical form.

• sin θ = 5/13,

  cos θ = 12/13, tan θ = 5/12, csc θ = 13/5, sec θ = 13/12, cot θ = 12/5

• sin θ = 12/13, cos θ = 5/13, tan θ = 12/5, csc θ = 13/12, sec θ = 13/5, cot θ = 5/12

• sin θ = 12/5,

  cos θ = 5/12, tan θ = 13/12, csc θ = 5/12, sec θ = 12/13, cot θ = 13/5

• sin θ = 5/12,

  cos θ = 12/5, tan θ = 5/13, csc θ = 12/5, sec θ = 5/13, cot θ = 12/13

Tags

CCSS.HSG.SRT.C.6

Find the missing length. Give your answer in simplest radical form.

• sin θ = 4/3,

  cos θ = 3/4,

  tan θ = √7/3, csc θ = √7/3, sec θ = 3/4,

  cot θ = 16/√7

• sin θ = 3/4,

  cos θ = √7/3, tan θ = 4/3,

  csc θ = 4/3,

  sec θ = 16/√7, cot θ = √7/3

• sin θ = √7/3, cos θ = 3/4,

  tan θ = √7/3, csc θ = 4√7/7, sec θ = 4/3,

  cot θ = 3√7/7

• sin θ = 3√7/7, cos θ = 4/3,

  tan θ = 3/4,

  csc θ = √7/3, sec θ = 16/√7, cot θ = √7/3

Tags

CCSS.HSG.SRT.C.6

Given the ratio for tan θ, find the remaining

ratios.

• sin𝛳=4/√41

  csc𝛳=5/√41

  cos𝛳=4/5

  sec𝛳=√41/4

  tan𝛳=4√41/5

  cot𝛳=5/4

• sin𝛳=5/√41

  csc𝛳=4/√41

  cos𝛳=3/5

  sec𝛳=√4/4

  tan𝛳=4√41/3

  cot𝛳=5/2

• sin𝛳=2/√41

  csc𝛳=5/√2

  cos𝛳=4/3

  sec𝛳=1/4

  tan𝛳=4/5

  cot𝛳=5/7

• sin𝛳=3/√41

  csc𝛳=2/√41

  cos𝛳=1/5

  sec𝛳=√41/7

  tan𝛳=4√41/5

  cot𝛳=5/2

Tags

CCSS.HSG.SRT.C.6

Given the ratio for csc θ, find the remaining ratios.

• sin𝛳=3/√7

  csc𝛳=√7/√7

  cos𝛳=1/√3

  sec𝛳=√5/2

  tan𝛳=√3/3

  cot𝛳=√3/7

• sin𝛳=2/√5

  csc𝛳=√5/√3

  cos𝛳=2/√7

  sec𝛳=√7/3

  tan𝛳=√7/√2

  cot𝛳=√7/2

• sin𝛳=2/√7

  csc𝛳=√3/√7

  cos𝛳=2/√3

  sec𝛳=√7/2

  tan𝛳=√7/√3

  cot𝛳=√3/2

• sin𝛳=1/√7

  csc𝛳=3/√7

  cos𝛳=2/3

  sec𝛳=√5/2

  tan𝛳=√5/√3

  cot𝛳=√2/2

Tags

CCSS.HSF.TF.C.8