Worksheets3rd Quarter Mastery Test in Elective Math (Feb 23, 2021)
Total questions: 50
Worksheet time: 25mins
Based on the given information determine the number of unique triangles that may exist.
B= 70°, b=85, c=88
1 Triangle
2 Triangles
3 Triangles
No Triangle
Based on the given information determine the number of unique triangles that may exist.
A= 37°, a=8, b=14
1 Triangle
2 Triangles
3 Triangles
No Triangle
Based on the given information determine the number of unique triangles that may exist.
J = 98°, j = 29, p = 6
1 Triangle
2 Triangles
3 Triangles
No Triangle
Based on the given information determine the number of unique triangles that may exist.
A= 137°, a=8, b=8
1 Triangle
2 Triangles
3 Triangles
No Triangle
Based on the given information determine the number of unique triangles that may exist.
A= 70°, c= 90, B = 58°
1 Triangle
2 Triangles
3 Triangles
No Triangle
Based on the given information determine the number of unique triangles that may exist.
E= 38.7°, f = 203, e = 172
1 Triangle
2 Triangles
3 Triangles
No Triangle
Based on the given information determine the number of unique triangles that may exist.
V = 45°, v = 83, a = 79
1 Triangle
2 Triangles
3 Triangles
No Triangle
Based on the given information determine the number of unique triangles that may exist.
d = 8, D = 49°, B = 57°
1 Triangle
2 Triangles
3 Triangles
No Triangle
Based on the given information determine the number of unique triangles that may exist.
K = 123°, k = 18, j = 12
1 Triangle
2 Triangles
3 Triangles
No Triangle
Based on the given information determine the number of unique triangles that may exist.
V = 29°, v = 12, w = 15
1 Triangle
2 Triangles
3 Triangles
No Triangle
Clint is building a wooden swing set for his children. Each supporting end of the swing set is to be an A-frame constructed with two 10 foot long 4-by-4s joined at a 45 degree angle. To prevent the swing set form tipping over, Clint wants to secure the base of each A-frame to concrete footings. How far apart should the footings for each A-frame be?
5.461 feet
6 feet
9.743 feet
7.65 feet
A triangle with two sides that measure 6 m and 8 m with an included angle of 137°.
17.8m²
14.6m²
16.4m²
12.9m²
A triangle with two sides that measure 5 cm and 8 cm with an included angle of 39°
15.9cm²
15.3cm²
12.6cm²
16.2cm²
Which of the following formulas shows the Law of Cosines?
c2 = a2 + b2 - 4ac + cosA
c2 = a2 - b2 - 2abcosC
c2 = a2 + b2 - 2abcosC
(cos A)/a = (cos B)/b
If three sides of a triangle are given, the Law of ____ is used to solve the triangle.
Sines
Cosines
Tangent
Cosecents
Find angle Z
20.15°
51.06°
57.71°
71.23°
Use Law of Cosines to find angle T
64.7°
62.2°
59.5°
53.1°
Find QR
34.7 km
31.1 km
13.74
2.2 km
Find the missing side.
521.1 in.
22.8 in.
19.2 in.
15.9 in.
Which Law would you use?
Law of Sines
Law of Cosines
Law of the Jungle
Law of Gravity
Find the area of ABC to the nearest tenth.
62.3 units2
77.0 units2
100 units2
124.6 units2
Solve each triangle. A=110, C=30, c=3
B=40, a=5.64, b= 3.86
B= 40, a= 7.90, b= 3.41
B= 45, a= 6.93, b= 2.76
B=40, a= 6.21, b= 3.86
Maruel must find the distance from Point A to Point B on opposite sides of a lake. He locates Point C that is 860 feet from point A and 175 feet from Point B. He measures the angle at Point C to be 78 degrees. Find the distance from point A to point B.
707643.58
877.62
842.01
841.22
Given: B = 70°, a = 11, C = 40° Find: c
70
11
8.5
7.5
Solve the Triangle
a=38, b= 31, c= 35
Find A, B and C
A= 43°, B= 47°, C= 90°
A= 50.1°, B= 70°, C= 59.9°
A= 70°, B= 50.1°, C= 59.9°
A= 60°, B= 70°, C= 50°
Find the area of ΔABC
if a = 9cm, b = 11cm, c = 16cm
89.5 sq cm
52.1 sq cm
47.6 sq cm
15.2 sq cm
In triangle ABC, a = 5, b = 6, c = 7. Which law would you need to use first?
Law of Sines
Law of Cosines
Both trig laws
Pythagorean Theorem
Find Area of ABC
223.6 units2
52.7 units2
73.1 units2
3.3 units2
Choose which law you need to solve the problem (DO NOT SOLVE):
Law of Sines
Law of Cosines
Both
Neither
In triangle ABC, a = 5, b = 6, c = 7. Which law would you need to use first?
Law of Sines
Law of Cosines
Both trig laws
Pythagorean Theorem
sin²θ
tanθ
cosθ
1- sin²θ
-1
1
csc θ
sin θ
1/cos x
cot x
1
-1
sin2 θ =
1 - cos2θ
sec2θ - 1
1 - sin2θ
csc2θ - 1
cot θ =
cosθ/sinθ
sinθ/cosθ
tanθ
1/cosθ
csc2 x
cot2 x
sec2 x
tan2 x
-(sin²x + cos²x)
-2
-1
1
2
sec θ =
1/cosθ
1/tanθ
1/sinθ
1/cscθ
Simplify the trig expression.
1
-1
-tanx
tanx
Simplify: sin x cot x
cos x
tan x
sin x
sec x
Simplify: sin x cot x
cos x
tan x
sin x
sec x
Simplify: tanxcotx-cos2x
tanx
cotx
sin2x
cos2x
Is this a pythagorean identity?
tan2x + 1 = secx
Yes
No
The expression sinA(cscA-sinA) is equivalent to
1
cos2A
-sin2A
cosA
sinx
tan²x
cot²x
cosx
-1
1
csc θ
sin θ
csc²θ
cscθ
sec²θ
1
csc²θ
cot θ tan θ
csc θ sec θ
1/sinθ
sinθ
cos²θ
sin²θ
1- sin²θ
csc x
1 + sec x
1/sec x
sec x
