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Lesson: Ambiguous Case of the Law of Sines

Video

10:43

Sample Question Videos

Worksheet • 10 Questions • 1 Video

Q1:

𝐴 𝐡 𝐢 is a triangle, where π‘š ∠ 𝐡 = 1 1 0 ∘ , 𝑏 = 1 6 cm, and 𝑐 = 1 2 cm. How many possible solutions are there for the other lengths and angles?

  • Atwo solutions
  • Bone solution
  • Cno solutions

Q2:

𝐴 𝐡 𝐢 is a triangle, where π‘š ∠ 𝐡 = 7 0 ∘ , 𝑏 = 3 cm, and 𝑐 = 6 cm. How many possible solutions are there for the other lengths and angles?

  • Atwo solutions
  • Bno solutions
  • Cone solution

Q3:

𝐴 𝐡 𝐢 is a triangle, where π‘š ∠ 𝐡 = 1 3 0 ∘ , 𝑏 = 1 7 cm, and 𝑐 = 3 cm. How many possible solutions are there for the other lengths and angles?

  • Atwo solutions
  • Bone solution
  • Cno solutions

Q4:

𝐴 𝐡 𝐢 is a triangle where π‘Ž = 1 3 . 8 c m , 𝑏 = 1 5 . 9 c m and π‘š ∠ 𝐴 = 2 8 ∘ . Find all possible values for the other lengths and angles giving lengths to two decimal places and angles to the nearest second.

  • A 𝑐 = 2 5 . 6 5 c m , π‘š ∠ 𝐡 = 3 2 4 4 β€² 4 5 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 1 9 1 5 β€² 1 5 β€² β€² ∘ or 𝑐 = 2 . 4 3 c m , π‘š ∠ 𝐡 = 1 4 7 1 5 β€² 1 5 β€² β€² ∘ , π‘š ∠ 𝐢 = 4 4 4 β€² 4 5 β€² β€² ∘
  • B 𝑐 = 2 5 . 6 5 c m , π‘š ∠ 𝐡 = 1 1 9 1 5 β€² 1 5 β€² β€² ∘ , π‘š ∠ 𝐢 = 3 2 4 4 β€² 4 5 β€² β€² ∘ or 𝑐 = 2 . 4 3 c m , π‘š ∠ 𝐡 = 4 4 4 β€² 4 5 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 4 7 1 5 β€² 1 5 β€² β€² ∘
  • C 𝑐 = 2 5 . 6 5 c m , π‘š ∠ 𝐡 = 3 2 4 4 β€² 4 5 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 1 9 1 5 β€² 1 5 β€² β€² ∘ or 𝑐 = 2 . 4 3 c m , π‘š ∠ 𝐡 = 4 4 4 β€² 4 5 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 4 7 1 5 β€² 1 5 β€² β€² ∘
  • D 𝑐 = 5 1 . 2 9 c m , π‘š ∠ 𝐡 = 3 2 4 4 β€² 4 5 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 1 9 1 5 β€² 1 5 β€² β€² ∘ or 𝑐 = 4 . 8 6 c m , π‘š ∠ 𝐡 = 1 4 7 1 5 β€² 1 5 β€² β€² ∘ , π‘š ∠ 𝐢 = 4 4 4 β€² 4 5 β€² β€² ∘

Q5:

𝐴 𝐡 𝐢 is a triangle where π‘Ž = 1 3 . 1 c m , 𝑏 = 3 0 . 3 c m and π‘š ∠ 𝐴 = 2 5 ∘ . Find all possible values for the other lengths and angles giving lengths to two decimal places and angles to the nearest second.

  • A 𝑐 = 3 0 . 2 2 c m , π‘š ∠ 𝐡 = 7 7 4 9 β€² 2 8 β€² β€² ∘ , π‘š ∠ 𝐢 = 7 7 1 0 β€² 3 2 β€² β€² ∘ or 𝑐 = 2 4 . 7 c m , π‘š ∠ 𝐡 = 1 0 2 1 0 β€² 3 2 β€² β€² ∘ , π‘š ∠ 𝐢 = 5 2 4 9 β€² 2 8 β€² β€² ∘
  • B 𝑐 = 3 0 . 2 2 c m , π‘š ∠ 𝐡 = 7 7 1 0 β€² 3 2 β€² β€² ∘ , π‘š ∠ 𝐢 = 7 7 4 9 β€² 2 8 β€² β€² ∘ or 𝑐 = 2 4 . 7 c m , π‘š ∠ 𝐡 = 5 2 4 9 β€² 2 8 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 0 2 1 0 β€² 3 2 β€² β€² ∘
  • C 𝑐 = 3 0 . 2 2 c m , π‘š ∠ 𝐡 = 7 7 4 9 β€² 2 8 β€² β€² ∘ , π‘š ∠ 𝐢 = 7 7 1 0 β€² 3 2 β€² β€² ∘ or 𝑐 = 2 4 . 7 c m , π‘š ∠ 𝐡 = 5 2 4 9 β€² 2 8 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 0 2 1 0 β€² 3 2 β€² β€² ∘
  • D 𝑐 = 6 0 . 4 5 c m , π‘š ∠ 𝐡 = 7 7 4 9 β€² 2 8 β€² β€² ∘ , π‘š ∠ 𝐢 = 7 7 1 0 β€² 3 2 β€² β€² ∘ or 𝑐 = 4 9 . 4 c m , π‘š ∠ 𝐡 = 1 0 2 1 0 β€² 3 2 β€² β€² ∘ , π‘š ∠ 𝐢 = 5 2 4 9 β€² 2 8 β€² β€² ∘

Q6:

𝐴 𝐡 𝐢 is a triangle where π‘š ∠ 𝐴 = 4 0 ∘ , π‘Ž = 5 c m and 𝑏 = 4 c m . If the triangle exists, find all the possible values for the other lengths and angles in 𝐴 𝐡 𝐢 giving lengths to two decimal places and angles to the nearest second.

  • A 𝑐 = 7 . 3 5 c m , π‘š ∠ 𝐡 = 3 0 5 6 β€² 4 6 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 0 9 3 β€² 1 4 β€² β€² ∘
  • BNo triangle exists.
  • C 𝑐 = 3 . 4 0 c m , π‘š ∠ 𝐡 = 3 0 5 6 β€² 4 6 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 0 9 3 β€² 1 4 β€² β€² ∘ or 𝑐 = 2 . 7 7 c m , π‘š ∠ 𝐡 = 1 4 9 3 β€² 1 4 β€² β€² ∘ , π‘š ∠ 𝐢 = 9 3 β€² 1 4 β€² β€² ∘
  • D 𝑐 = 3 . 4 0 c m , π‘š ∠ 𝐡 = 3 0 5 6 β€² 4 6 β€² β€² ∘ , π‘š ∠ 𝐢 = 1 0 9 3 β€² 1 4 β€² β€² ∘
  • E 𝑐 = 7 . 3 5 c m , π‘š ∠ 𝐡 = 1 0 9 3 β€² 1 4 β€² β€² ∘ , π‘š ∠ 𝐢 = 3 0 5 6 β€² 4 6 β€² β€² ∘

Q7:

𝐴 𝐡 𝐢 is a triangle, where π‘Ž = 2 8 c m , 𝑏 = 1 7 c m , and π‘š ∠ 𝐢 = 6 0 ∘ . Find the missing length rounded to three decimal places and the missing angles rounded to the nearest degree.

  • A 𝑐 = 2 4 . 4 3 4 c m , π‘š ∠ 𝐴 = 8 3 ∘ , π‘š ∠ 𝐡 = 3 7 ∘
  • B 𝑐 = 3 0 . 8 8 7 c m , π‘š ∠ 𝐴 = 6 4 ∘ , π‘š ∠ 𝐡 = 5 6 ∘
  • C 𝑐 = 1 5 . 7 6 5 c m , π‘š ∠ 𝐴 = 1 1 7 ∘ , π‘š ∠ 𝐡 = 3 ∘
  • D 𝑐 = 2 8 . 8 9 6 c m , π‘š ∠ 𝐴 = 7 0 ∘ , π‘š ∠ 𝐡 = 5 0 ∘

Q8:

𝐴 𝐡 𝐢 is a triangle, where π‘Ž = 2 8 c m , 𝑏 = 2 4 c m , and π‘š ∠ 𝐢 = 4 0 ∘ . Find the missing length rounded to three decimal places and the missing angles rounded to the nearest degree.

  • A 𝑐 = 1 8 . 1 7 8 c m , π‘š ∠ 𝐴 = 8 2 ∘ , π‘š ∠ 𝐡 = 5 8 ∘
  • B 𝑐 = 3 3 . 2 0 6 c m , π‘š ∠ 𝐴 = 5 6 ∘ , π‘š ∠ 𝐡 = 8 4 ∘
  • C 𝑐 = 2 2 . 2 7 3 c m , π‘š ∠ 𝐴 = 7 4 ∘ , π‘š ∠ 𝐡 = 6 6 ∘
  • D 𝑐 = 2 9 . 0 7 3 c m , π‘š ∠ 𝐴 = 6 3 ∘ , π‘š ∠ 𝐡 = 7 7 ∘

Q9:

𝐴 𝐡 𝐢 is a triangle, where π‘š ∠ 𝐴 = 7 0 ∘ , 𝐡 𝐢 = 3 c m , and 𝐴 𝐢 = 3 9 c m . If the triangle exists, find all the possible values for the other lengths and angles in β–³ 𝐴 𝐡 𝐢 giving the lengths to two decimal places and angles to the nearest degree.

  • A 𝐴 𝐡 = 𝐴 𝐡 = 3 6 . 6 5 c m , π‘š ∠ 𝐡 = 9 0 ∘ , π‘š ∠ 𝐢 = 2 0 ∘
  • BThe triangle does not exist.
  • C 𝐴 𝐡 = 𝐴 𝐡 = 3 8 . 8 8 c m , π‘š ∠ 𝐡 = 9 0 ∘ , π‘š ∠ 𝐢 = 2 0 ∘

Q10:

𝐴 𝐡 𝐢 is a triangle, where π‘š ∠ 𝐴 = 2 0 ∘ , 𝐡 𝐢 = 3 c m , and 𝐴 𝐢 = 3 0 c m . If the triangle exists, find all the possible values for the other lengths and angles in β–³ 𝐴 𝐡 𝐢 giving the lengths to two decimal places and angles to the nearest degree.

  • A 𝐴 𝐡 = 𝐴 𝐡 = 1 0 . 2 6 c m , π‘š ∠ 𝐡 = 9 0 ∘ , π‘š ∠ 𝐢 = 7 0 ∘
  • BThe triangle does not exist.
  • C 𝐴 𝐡 = 𝐴 𝐡 = 2 9 . 8 5 c m , π‘š ∠ 𝐡 = 9 0 ∘ , π‘š ∠ 𝐢 = 7 0 ∘
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