AnswersA4142 Geometry 2Test

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Triangle ABC is a right triangle and cos(22.6o)=. Solve for b and round to the nearest whole number.

Question illustration
A
tan(22.6o) =
Option A
B
tan(22.6o) =
Option B
C
tan(22.6o) =
Option C
D
tan(22.6o) =
Option D
2
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A right triangle has one angle that measure 23o. The adjacent leg measures 27.6 cm and the hypotenuse measures 30 cm.What is the approximate area of the triangle? Round to the nearest tenth.Area of a triangle = bh

Question illustration
A
68.7 cm2
B
161.8 cm2
C
381.3 cm2
D
450.0 cm2
3

Law of sines:

Question illustration
A
only 90°
B
only 155°
C
20° and 110°
D
45° and 135°
4

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

Question illustration
A
92 = y2 + 192 – 2(y)(19)cos(41°)
B
y2 = 92 + 192 – 2(y)(19)cos(41°)
C
92 = y2 + 192 – 2(9)(19)cos(41°)
D
y2 = 92 + 192 – 2(9)(19)cos(41°)
5

Two teams are pulling a heavy chest, located at point X. The teams are 4.6 meters away from each other. Team A is 2.4 meters away from the chest, and Team B is 3.2 meters away. Their ropes are attached at an angle of 110°.Law of sines:

Question illustration
A
StartFraction sine (uppercase A) Over 2.4 EndFraction = StartFraction sine (110 degrees) Over 4.6 EndFraction
Option A
B
StartFraction sine (uppercase A) Over 4.6 EndFraction = StartFraction sine (110 degrees) Over 2.4 EndFraction
Option B
C
StartFraction sine (uppercase A) Over 3.2 EndFraction = StartFraction sine (110 degrees) Over 4.6 EndFraction
Option C
D
StartFraction sine (uppercase A) Over 4.6 EndFraction = StartFraction sine (110 degrees) Over 3.2 EndFraction
Option D
6

In right triangle XYZ, the right angle is located at vertex Y. The length of line segment XY is 12.4 cm. The length of line segment YZ is 15.1 cm.

A
34.8°
B
39.4°
C
50.6°
D
55.2°
7

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

Question illustration
A
35°
B
45°
C
55°
D
65°
8

In which triangle is the value of x equal to tan−1? (Images may not be drawn to scale.)

Question illustration
A
A right triangle is shown. The length of the hypotenuse is 5.2 and the length of the side adjacent to the right angle is 3.1. The angle between the 2 sides is x.
Option A
B
A right triangle is shown. The length of the hypotenuse is 5.2 and the length of the side adjacent to the right angle is 3.1. The angle opposite to side with length 3.1 is x.
Option B
C
A right triangle is shown. The length of 2 sides are 5.2 and 3.1. The angle opposite to side with length 5.2 is x.
Option C
D
A right triangle is shown. The length of 2 sides are 5.2 and 3.1. The angle opposite to side with length 3.1 is x.
Option D
9

Which measures are accurate regarding triangle JKL? Select two options.

A
m∠K = 94°
B
k ≈ 3.7 units
C
k ≈ 4.6 units
D
KL ≈ 2.5 units
E
KL ≈ 3.2 units
10

Triangle L J K is shown. Angle J L K is a right angle. The length of the hypotenuse is 15 inches and the length of the side adjacent to the right angle is 10 inches. The angle between the 2 sides is x.

Question illustration
A
sin(x) =
Option A
B
sin(x) =
Option B
C
cos(x) =
Option C
D
cos(x) =
Option D
11

The equation can be used to find the measure of angle LKJ.

Question illustration
A
41°
B
45°
C
49°
D
55°
12

The law of cosines for RST can be set up as 52 = 72 + 32 – 2(7)(3)cos(S). What could be true about RST?Law of cosines: a2 = b2 + c2 – 2bccos(A)

Question illustration
A
r = 5 and t = 7
B
r = 3 and t = 3
C
s = 7 and t = 5
D
s = 5 and t = 3
13

Law of sines:

Question illustration
A
15°
B
30°
C
45°
D
60°
14

Which equation can be used to find the length of ?

Question illustration
A
(10)sin(40o) = AC
B
(10)cos(40o) = AC
C
= AC
Option C
D
= AC
Option D
15

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

Question illustration
A
34°
B
41°
C
51°
D
56°
16

The equation can be used to find the length of .

Question illustration
A
19.3 in.
B
21.3 in.
C
23.5 in.
D
68.0 in.
17

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

Question illustration
A
10
B
13
C
18
D
21
18

Right triangle ABC is shown.

Question illustration
A
sin(50o) =
Option A
B
sin(50o) =
Option B
C
cos(50o) =
Option C
D
cos(50o) =
Option D
19

Law of sines:

Question illustration
A
20°
B
34°
C
41°
D
53°
20

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

Question illustration
A
62 = p2 + 82 – 2(p)(8)cos(39°)
B
p2 = 62 + 82 – 2(6)(8)cos(39°)
C
82 = 62 + p2 – 2(6)(p)cos(39°)
D
p2 = 62 + 62 – 2(6)(6)cos(39°)

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