MATH 118
Solutions: TEST III
Instructions: For each of the following problems be certain to show your work!
Answer any 15 of the following 18 questions. You may answer more than 18 to obtain extra
credit.
1.
Find the area of triangle UVW. Express your answer to the nearest tenth. Assume that the units are meters.
Solution:
Strategy: using the law of sines we will find the length of UW; then we will apply the area formula, A = ½ ab sin C.
First note that angle V = 180 – 28 – 37 = 115
Next:
UW/ sin 115 = 25/sin 37
So UW = (25 sin 115)/ sin 37 ≈ 37.648
Finally, the area of triangle UVW = ½ (37.648)(25)sin 28 ≈ 220.9 square meters.
2.
Find y. Express your answer to the nearest tenth. Assume that the units are meters.
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Solution: Here we employ the law of cosines, viz.
y2 = 52 + 72 – 2 (5)(7)cos 83 = 74 – 70 cos 83 ≈ 65.469
Hence 𝑦 = √65.469 ≈ 𝟖. 𝟏 𝒎𝒆𝒕𝒆𝒓𝒔
3. (𝑎)
𝜃 is an acute angle. Find the value of 𝜃 in degrees if sec 𝜃 =
1
3
Solution: 𝑐𝑜𝑠 𝜃 = 𝑠𝑒𝑐 𝜃 = 2√3 =
(𝑏)
√3
.
2
2√3
3
.
𝑇ℎ𝑢𝑠, 𝑠𝑖𝑛𝑐𝑒 𝜃 𝑖𝑠 𝑎𝑐𝑢𝑡𝑒, 𝜽 = 𝟑𝟎°
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Assume that 0 ≤  ≤ 2. Find all solutions to the equation cos  = 3.
Solution: Since cos x ≤ 1 for all x, there is no solution to the given equation.
(c)
Find tan(arc cos(4/5)) without using a calculator.
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Solution: 𝐿𝑒𝑡 𝜃 = 𝑎𝑟𝑐 𝑐𝑜𝑠 5 . 𝑇ℎ𝑒𝑛 𝑐𝑜𝑠 𝜃 = 5 . So tan 𝜃 = 3/4
4.
Find y and z. Assume that the units are feet.
𝑦/ sin 120° = 3/ sin 16°.
Solution: Using the law of sines:
Hence
𝑦=
3 sin 120°
sin 16°
≈ 𝟗. 𝟒 𝒇𝒆𝒆𝒕
We will also apply the law of sines to find z. Begin by noting that angle Z = 𝑎𝑛𝑔𝑙𝑒 𝑍 = 180° − 16° −
120° = 44°. 𝑁𝑜𝑤
Hence 𝑧 =
3 sin 44°
sin 16°
𝑧
sin 44°
=
3
sin 16°
.
≈ 𝟕. 𝟔 𝒇𝒆𝒆𝒕
5. (a) What is the amplitude of a sine function that has a maximum of 11 and a minimum of 5?
Solution: Since peak to peak is 11 – 5 = 6, the amplitude is ½ (6) = 3.
(b) What is the period of the function y = 9 + 13 sin(5x)?
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Solution: The period = (2)/(5) = 2/5
6.
Even, odd, or neither? Explain each answer.
(a) y = x2 + x3 + 4
Neither, since f(1) = 6 and f(-1) = 4.
(b) y = x2 cos3 x
This is a product of even functions; hence even.
(c)
y = x3 sin x
This is a product of two odd functions; hence even.
(d) y = x2 + sin(x/2)
This is a sum of an even function and an odd function. We expect that it is neither even nor odd. Let’s check:
When x = 1, y = 2.
When x = -1, y = 0.
Thus this function is neither even nor odd.
7. The regular dodecahedron, often simply called "the" dodecahedron, is the Platonic solid composed of 20 polyhedron
vertices, 30 polyhedron edges, and 12 pentagonal faces, as pictured below. If the each edge is 8 cm long, find its
total surface area.
Solution: Since there are 12 pentagonal faces, we should find the area of one pentagonal face which we
then multiply by 12.
Each face consists of 5 isosceles triangles, each with internal angle of ½ (72) = 36 and base length of 8
cm. The other two angles are 54 and 54.
Looking at the sketch below, we find c = 8 and a = 4 tan 54.
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Since the area of such a triangle is ½ (8) (4 tan 54) = 16 tan 54, we find that the surface area of the 16
tan 54 is 12 (16 tan 54) = 192 tan 54 ≈ 264.3 cm2.
8. Find the length of the altitude WY.
Solution: Let h = YW. Then sin 32° = h/6. From this it follows that h = 6 sin 32° ≈ 3.18
9. What is the exact time between 6 and 7 o’clock when the two hands of a clock are pointing in the same direction?
Solution: Let the exact time be 6 hours and x minutes. Then, when the hour hand advances by 30 and x
minutes, the minute hand will have traveled 1/12 (30 + x) minutes. Hence
x = 1/12 (30 + x) → 12x = 30 + x →x = 30/11 = 2 and 8/11.
Thus the exact time will be 6:32 and 8/11, or approximately 6:32 minutes and 43.6 seconds.
10. (a) Evaluate cos 5 cos 175 – sin 5 sin 175 without the use of a calculator. Show your work and give
an exact answer.
Solution: cos 5 cos 175 – sin 5 sin 175 = cos(5 + 175) = cos 180 = -1
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(b) Given that sin x = 8/17 and cos y = 3/5, find sin(x + y) without the use of a calculator. Show your
work and give an exact answer.
Solution: sin (x + y) = sin x cos y + cos x sin y = (8/17) (3/5) + (15/17)(4/5) = 84/85.
11. Which of the following are right triangles? Explain your answer for each one.
(a) 16, 63, 65
Observe that 162 + 632 = 256 + 3969 = 4225
and 652 = 4225
Thus this is a right triangle.
(b) 19, 180, 182
Observe that 192 + 1802 = 361 + 32400 = 32761 ≠ 33124 = 1822
Thus this is not a right triangle.
(c)
800, 1500, 1700
Observe that 8002 + 15002 = 640000 + 2250000 = 2890000
and 17002 = 2890000
Thus this is a right triangle.
12. Find the smallest angle of a triangle with sides of length 49, 52 and 70 feet. (Give your answer to the
nearest tenth of a degree.)
Solution: The smallest angle is the one opposite the shortest side; in this case 49.
Using the law of cosines,
c2 = a2 + b2 – 2 ab cos→ 𝜃 = arccos
Hence = 44.4°
𝑎2 +𝑏 2 −𝑐 2
2𝑎𝑏
= arccos
522 +702 −492
2(52)(70)
=
6
13. Albertine, a surveyor, wants to measure the width of a river. She stands at a point A on the bank of the
river, with point B opposite her on the other side of the river. From A she walks 350 feet along the bank to a
point C. The line CA makes an angle of 41 with the line CB. Find the width of the river to the nearest tenth of
a foot.
Solution:
A
350 ft
41
C
B
The width of the river, AB, equals 350 tan 41 ≈ 304.3 ft.
14. Swann has inherited a commercial triangular-shaped lot in Nice of side lengths 413 meters, 213 meters, and
299 meters. The neighboring property is selling for one million euros per hectare. How much is Swann’s
property worth? (Hint: A hectare is the equivalent of 10,000 square meters.)
Solution:
Let a = 213, b = 413, and c = 299.
Using the law of cosines,
c2 = a2 + b2 – 2 ab cosC→
𝐶 = arccos
𝑎2 +𝑏2 −𝑐 2
2𝑎𝑏
= arccos
2132 +4132 −2992
2(213)(413)
= arccos( 0.719214) = 44.0
Thus the area of triangle ABC is ½ ab sin C = ½ (213)(413)sin(44.0 ) =30554.2 m2 = 3.05542 hectares.
This may be worth about 3.06 million dollars if his reasoning is correct.
15. If 4 sin2 x – 8 sin x + 3 = 0, solve for x subject to the constraint that 0 ≤ x ≤ /2.
Solution: Let z = sin x. Then 4 z2 – 8 z + 3 = 0. Factoring: (2x – 3)(2z – 1) = 0.
Hence z = 3/2 or z = ½.
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So we must solve sin x = 3/2 and sin x = ½. The former has no solution since sin x ≤ 1 for all x.
The latter has the solution x = /6.
16. The Eiffel tour, completed in 1889, is 324 meters high. Albertine is standing due west of the tower;
Bernadette is standing due north of the tower. Albertine’s angle of elevation is 1.27 radians;
Bernadette’s is 1.02 radians. How far apart are Albertine and Bernadette?
Solution: Let T be the point at which the tower touches the ground. Let A be the point at which Albertine
stands and let B be the point where Bernadette stands. Let a = distance from A to T and let b = distance
from B to T. Now a = 324 cot 1.27 and b = 324 cot 1.02.
B
A
T
Now, since ABT is a right triangle, the distance from A to B is
√𝑎2 + 𝑏 2 = √(324 cot 1.27) 2 + (324 cot 1.02)2 = 324√(cot 1.27) 2 + (cot 1.02)2 ≈ 𝟐𝟐𝟐. 𝟗 𝒎𝒆𝒕𝒆𝒓𝒔
17. If 3 cot ∝ = 2, find the value of
8
10 sin ∝ −6 cos ∝
4 sin ∝ +3 cos ∝
Express your answer to the nearest tenth.
Solution: First observe that, since cot ∝ =2/3, tan ∝ = 3/2. 𝑇ℎ𝑢𝑠
1
10 sin ∝ −6 cos ∝ 10 sin ∝ −6 cos ∝ cos ∝
=
∙
=
1
4 sin ∝ +3 cos ∝
4 sin ∝ +3 cos ∝
cos ∝
2
sin ∝
10 ( ) − 6 20 − 18
10
−6
10
tan
𝛼
−
6
3
cos ∝
=
=
=
= −2.0
sin ∝
2
4
tan
𝛼
+3
8
−
9
4
+3
4( ) +3
cos ∝
3
18. From the top of a cliff 150 feet high, the angles of depression of two boats which are due South of the
observer are 17 and 80. Find the distance that separates the two boats. Express your answer to the nearest
tenth of a foot.
Solution:
17 deg
80 deg
150 ft
Let x = distance from base of cliff to the nearer boat.
Let y = distance from base of cliff to the more distant boat.
Then the distance between the two boats is y – x = 150 cot 17 – 150 cot 80 = 150 (cot 17 – cot 80) = 464.2 feet
The universe stands continually open to our gaze, but it cannot be understood unless one first
learns to comprehend the language and interpret the characters in which it is written. It is
written in the language of mathematics, and its characters are triangles, circles, and other
geometric figures, without which it is humanly impossible to understand a single word of it;
without these, one is wandering about in a dark labyrinth.
- Galileo Galilei
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