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Author: nsharp1
No. Questions: 10
Created: Feb 17, 2021
Last Modified: 4 years ago

Prove Law of Sines, Acute

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The questions below concern the following proof.


Let ΔABC be an acute triangle. Let A represent mA, B represent mB, and C represent mC. Prove the law of sines, BCsin(A)=ACsin(B)=ABsin(C).

Acute Triangle ABC v2

Statement Reason
1.Construct altitude ¯BD such that D lies on ¯AC1.
2.¯BD¯AC2.Definition of an altitude
3.ADB, BDC are right angles3.Definition of perpendicular lines
4.ΔADB, ΔBDC are right triangles4.Definition of right triangles
5.5.Sine ratio in a right triangle
6.ABsin(A)=BD6.Multiplication Property of Equality
7.7.Sine ratio in a right triangle
8.BCsin(C)=BD8.Multiplication Property of Equality
9.ABsin(A)=BCsin(C)9.
10.AB=BCsin(C)sin(A)10.Division Property of Equality
11.ABsin(C)=BCsin(A)11.Division Property of Equality
12.Construct altitude ¯AE such that E lies on ¯BC12.
13.¯AE¯BC13.Definition of an altitude
14.AEB, AEC are right angles14.Definition of perpendicular lines
15.ΔAEB, ΔAEC are right triangles15.Definition of right triangles
16.16.Sine ratio in a right triangle
17.ACsin(C)=AE17.Multiplication Property of Equality
18.18.Sine ratio in a right triangle
19.ABsin(B)=AE19.Multiplication Property of Equality
20.20.Transitive Property of Equality
21.ABsin(B)sin(C)=AC21.Division Property of Equality
22.ABsin(C)=ACsin(B)22.Division Property of Equality
23.BCsin(A)=ACsin(B)23.
24.BCsin(A)=ACsin(B)=ABsin(C)24.Combined results
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
A.
What is the missing reason in step 1?
  1. Given
  2. All acute triangles have altitudes from each vertex that intersect the opposite side
  3. Perpendicular Bisector Theorem (for triangles)
  4. All acute triangles have an orthocenter that lies within the triangle
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
B.
What is the missing statement in step 5?
  1. sin(A)=BDAB
  2. sin(A)=BDAD
  3. sin(A)=BCAC
  4. sin(A)=ADAB
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
C.
What is the missing statement in step 7?
  1. sin(C)=BDAB
  2. sin(C)=CSBC
  3. sin(C)=BCAC
  4. sin(C)=BDBC
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
D.
What is the missing statement in step 9?
  1. Transitive Property of Equality
  2. Law of sines
  3. Multiplication Property of Equality
  4. Trigonometric Identity
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
E.
What is the missing reason in step 12?
  1. Each of the three altitudes of an acute triangle intersects the side opposite the vertex through which the altitude passes
  2. The orthocenter of all acute triangles lies within the triangle
  3. Perpendicular Bisector Theorem (for triangles)
  4. All triangles have three altitudes
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
F.
What is the missing statement in step 16?
  1. sin(C)=ABAC
  2. sin(C)=CEAC
  3. sin(C)=AEAC
  4. sin(C)=AECE
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
G.
What is the missing statement in step 18?
  1. sin(B)=CDBC
  2. sin(B)=BEAB
  3. sin(B)=ABAC
  4. sin(B)=AEAB
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
H.
What is the missing statement in step 20?
  1. ABsin(B)=ACsin(C)
  2. AE=AE
  3. sin(C)=sin(B)
  4. AC=AB
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
I.
What is the missing reason in step 23?
  1. Previous result
  2. Transitive Property of Equality
  3. Multiplication Property of Equality
  4. Law of sines
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
J.
The proof given applies to which type of triangles?
  1. All triangles.
  2. Acute triangles.
  3. Non-right triangles.
  4. Scalene triangles.