AnswersGeometry - Semester 1 PathwaysTriangle Congruence: SSS and HL

Triangle Congruence: SSS and HL — Unit test Answers

15 verified answers
4

The equation can be used to find the length of .

Question illustration
A
3.0 cm.
B
9.8 cm.
C
10.5 cm.
D
12.8 cm.
7

Law of cosines: a2 = b2 + c2 – 2bccos(A)

Question illustration
A
72 = 82 + 112 – 2(8)(11)cos(N)
B
82 = 72 + 112 – 2(7)(11)cos(M)
C
72 = 82 + 112 – 2(8)(11)cos(P)
D
82 = 72 + 112 – 2(7)(11)cos(P)
9

The equation can be used to find the length of .

Question illustration
1
14.3 in.
2
20.5 in.
2
21.3 in.
2
22.6 in.
12

Triangle A B C is shown. The length of A B is 12, the length of B C is 24, and the length of C A is 12 StartRoot 3 EndRoot

Question illustration
A
m∠A = 30°, m∠B = 60°, m∠C = 90°
B
m∠A = 90°, m∠B = 60°, m∠C = 30°
C
m∠A = 60°, m∠B = 90°, m∠C = 30°
D
m∠A = 90°, m∠B = 30°, m∠C = 60°
14

Law of sines:

Question illustration
A
StartFraction sine (51 degrees) Over 2.6 EndFraction = StartFraction sine (76 degrees) Over z EndFraction
Option A
B
StartFraction sine (51 degrees) Over 2.6 EndFraction = StartFraction sine (53 degrees) Over z EndFraction
Option B
C
StartFraction sine (76 degrees) Over 2.6 EndFraction = StartFraction sine (51 degrees) Over z EndFraction
Option C
D
StartFraction sine (76 degrees) Over 2.6 EndFraction = StartFraction sine (53 degrees) Over z EndFraction
Option D

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