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Trigonometry Unit Practice Test Solutions
- Add degrees, minutes, and seconds separately:
0
0
0
Since 60’ form 1 degree we will leave 10’ in the minute’s place and add a degree to 113 degrees ( 70 ᇱ^ ൌ 60ᇱ^ 10ᇱ^ ൌ 1°10Ԣ). So, the answer is °Ԣ.
- We will need to rewrite 45°05ᇱ^ as 44°64Ԣ60" to ensure that we are able to carry out simple subtraction in each category. (1 degree is equal to 60 minutes and one minute is equal to 60 seconds, so 1° ൌ 59Ԣ60"ሻ.
0
0
0
The answer is °Ԣ".
- 30°15Ԣ06" ൌ 30° 15 ൈ ቀ^ ଵቁ ° 6 ൈ ቀ^ ଷଵ ቁ ° ൌ 30° 0.25° 0.0017° ൌ 30.2517° Here we used the fact that 1’= (^) ଵ degree, and 1" ൌ ቀ (^) ଵቁ
ᇱ ൌ (^) ଵ ൈ ቀ (^) ଵቁ ° ൌ ቀ (^) ଷଵ ቁ °. The answer is . ૠ°.
4. 12.3456° ൌ 12° 0.3456 ൈ 1° ൌ 12° 0.3456 ൈ 60ᇱ^ ൌ 12° 20.736ᇱ^ ൌ 12° 20ᇱ^
0.736 ൈ 1ᇱ^ ൎ 12° 20ᇱ^ 44"
We replaced 1° with 60Ԣ and 1’ with 60” in our calculations and rounded last multiplication of 0.736 ൈ 60" ൌ 44.16" ൎ 44" to the nearest second as directed.
The answer is °Ԣ".
Trigonometry Unit Practice Test Solutions
Set calculator mode to degrees!
5. sin 15 010 '= sin 15. 16670 ≈ 0. 2616.
The answer is 0.
6. cos 80012 "= cos 80. 00330 ≈ 0. 1736. The answer is 0.
7. tan 48015 '= tan 48. 250 ≈ 1. 1204.
The answer is 1.
sin 8. 0042
csc 8015 "= csc 8. 00420 = 0 ≈.
Use: θ
θ
sin
csc =.
The answer is 7.
cos 80. 0033
sec 8012 " sec 80. 0033 1
0
Use: θ
θ
cos
sec = 1.
The answer is 5.
tan 85. 1033
cot 850612 ' " cot 85. 1033 1
0
Use: θ
θ
tan
cot = 1.
The answer is 0.
- cosି ଵ^ ሺ0.9876ሻ ൎૢ. ° or 9°01'56"
- sinିଵ^ ሺ0.5602ሻ ൎ . ૠ° or 34°04'05"
- tanି ଵ^ ሺ5ሻ ൎ ૠૡ. ૢ° or 78°41'24"
- Calculators do not have a button with cscି ^. We will use the fact that for positive angle values
within zero to 90 degree range ⎟
θ
θ
csc 1 sin^1.
csc 25 sin ⎟≈
− = −⎛ or °ૠԢ"
Trigonometry Unit Practice Test Solutions
- To solve a triangle is to find all missing sides and angles. cos 39.1°ൌ (^) ଵହ
bൌ150ൈcos 39.1° ൎ 116.
We rounded the length of side AC to the nearest whole number since side AB is given with precision of a whole number.
Similarly,
sin39.1° ൌ (^) ଵହ
ܽൌ 150 ൈ sin 39.1° ൎ 95.
Interior angles of a triangle add to ૡ°
39.1° ܤס ݉ 90° ൌ 180° ݉ ܤס 129.1° ൌ 180° ݉ ܤסൌ 50.9°
Solution:
tan ݔൌ
ݔൌ tanି ଵ^ ൬
555 ft
444 ft
Angle of elevation
x
C a B
b
Trigonometry Unit Practice Test Solutions
- South‐West direction implied the third quadrant. Note that angles with compass directions are measured from the vertical, North‐South, line!
cos 30.5° ൌ
ݔൌ 200 ൈ cos 30.5° ൎ 172 miles
Apply Law of Sines to solve for a:
sin 40°
sin 68°ܽ
0.9 ൈ sin 68° sin 40° ൎ . miles Apply Law of Sines to solve for b: sin 40°
sin 72°ܾ
0.9 ൈ sin 72° sin 40° ൎ . miles
The eagle is 1.33 miles away from lookout A and 1.30 miles away from lookout B.
0. 9 mi
A
B
b
a
C
ship
South
West
200 mi
x
Trigonometry Unit Practice Test Solutions
- This problem is solved with the Law of Cosines because lengths of all three sides are known.
400 ଶ^ ൌ 350 ଶ^ 200ଶ^ െ 2 ൈ 350 ൈ 200 ൈ cos ܣ
160,000 ൌ 125,500 40,000 െ 140,000 cos ܣ
160,000 ൌ 165,500 െ 140,000 cos ܣ
െ5,500 ൌ െ140,000 cos ܣ
cos ܣൌ
݉ ܣסൌ cosି ଵ^ ൬
Note: angle A must be the largest in the triangle since it is opposite the longest side.
You can use either of the laws (sines or cosines) to find one of the remaining angles. ݉ ܤסൌ . °. Then use the fact that the sum of interior angles of a tringle is equal to 180 degrees to solve for angle C. ݉ ܥסൌ 180° െ 87.75° െ 61.03° ൌ . °.
- This problem is solved with the Law of Sines.
0
0
sin 0. 6666173927 41. 8
80. 0 sin 105. 4
sin
sin
sin 105. 4
×
m A −
A
A
Supplement of angle A, 138.2°, also has the same value of sine, but we already have an obtuse angle in the triangle and a triangle cannot have more than one!
(solution continues on the next page)
A
B
C
A
B
C
Trigonometry Unit Practice Test Solutions
sin 105.4° 115.7 ൌ
sin 32.8°ܾ
115.7 ൈ sin 32.8° sin 105.4° ൎ 65.
The solution is the unique triangle: ܽ ൌ 80.0 ܣൌ 41.8°ܾ ൌ 65.0 ܤൌ 32.8°ܿ ൌ 115.7 ܥൌ 105.4°
sin 40°
sin ܤ
sin ܤൌ 5.6 ൈ sin 40°
No angle can satisfy the requirement of value of sine to be greater than 1!
No solution!
Look at the diagram below, side CB is too short for a 40 ‐degree angle with a side of 5.6 units!
A B
C
A
B
C
gap!
Trigonometry Unit Practice Test Solutions
- This problem can be solved with Law of Cosines:
ݔ ଶ^ ൌ 300 ଶ^ 300ଶ^ െ 2ሺ300ሻ ሺ300ሻcos 120.25°
ݔ ଶ^ ൎ 270,679.
ݔൌ ඥ270,679.3159 ൎ miles
The ships are approximately 520 miles apart.
- Use Heron’s formula, ܽെ ݏሺݏඥ ൌ ܣ ܾെ ݏሻሺ ܿെ ݏሻሺ ሻ, where s is the semiperimeter. ൌ ݏ
ܣൎ 196,388.2873 sq.ft 196,388. 43, ൎ 4.5 acres There are 4.5 acreas in the triangular field.
- Area: ൌ ܣ ଵଶ ൈ 55 ൈ 40 ൈ sin 95° ൎ 1,096 sq.ft
450 ft
900 ft
1100 ft
H^ ( harbor^ )
A
B
300 mi
x
40 ft
55 ft
