D
OB, so we need to find OB. To do this, we will show that △ OBC
37◦
is a right triangle, then find the angle ∠ BOC, and then find BC.
B
C
The length OB will then be simple to determine.
8
P
.3
1
Since the slanted sides are tangent to each roller, ∠ OD A =
E
∠ PEC = 90◦. By symmetry, since the vertical line through the
centers of the rollers makes a 37◦ angle with each slanted side,
A
we have ∠ OAD = 37◦. Hence, since △ OD A is a right triangle,
∠ DOA is the complement of ∠ OAD. So ∠ DOA = 53◦.
Since the horizontal line segment BC is tangent to each roller, ∠ OBC = ∠ PBC = 90◦. Thus,
△ OBC is a right triangle. And since ∠ OD A = 90◦, we know that △ ODC is a right triangle. Now,
OB = OD (since they each equal the radius of the large roller), so by the Pythagorean Theorem we
have BC = DC:
BC 2 = OC 2 − OB 2 = OC 2 − OD 2 = DC 2
⇒
BC = DC
Thus, △ OBC and △ ODC are congruent triangles (which we denote by △ OBC ∼
= △ ODC), since
their corresponding sides are equal. Thus, their corresponding angles are equal. So in particular,
∠ BOC = ∠ DOC. We know that ∠ DOB = ∠ DOA = 53◦. Thus,
53◦ = ∠ DOB = ∠ BOC + ∠ DOC = ∠ BOC + ∠ BOC = 2 ∠ BOC
⇒
∠ BOC = 26.5◦ .
Likewise, since BP = EP and ∠ PBC = ∠ PEC = 90◦, △ BPC and △ EPC are congruent right triangles. Thus, BC = EC. But we know that BC = DC, and we see from the diagram that EC+ DC = 1.38.
Thus, BC + BC = 1.38 and so BC = 0.69. We now have all we need to find OB:
BC
BC
0.69
= tan ∠ BOC
⇒
OB =
=
= 1.384
OB
tan ∠ BOC
tan 26.5◦
Hence, the diameter of the large roller is d = 2 × OB = 2(1.384) = 2.768 .
5This will often be worded as the line that is tangent to the circle.
18
Chapter 1 • Right Triangle Trigonometry
§1.3
Example 1.18
A slider-crank mechanism is shown in Figure 1.3.2 below. As the piston moves downward the con-
necting rod rotates the crank in the clockwise direction, as indicated.
piston
A
a
C
conne b
c
c
ting
rod
B
θ
crank
r
O
Figure 1.3.2
Slider-crank mechanism
The point A is the center of the connecting rod’s wrist pin and only moves vertically. The point B
is the center of the crank pin and moves around a circle of radius r centered at the point O, which
is directly below A and does not move. As the crank rotates it makes an angle θ with the line O A.
The instantaneous center of rotation of the connecting rod at a given time is the point C where the
horizontal line through A intersects the extended line through O and B. From Figure 1.3.2 we see that ∠ OAC = 90◦, and we let a = AC, b = AB, and c = BC. In the exercises you will show that for 0◦ < θ < 90◦,
b 2 − r 2 (sin θ)2
c =
and
a = r sin θ +
b 2 − r 2 (sin θ)2 tan θ .
cos θ
Applications and Solving Right Triangles • Section 1.3
19
For some problems it may help to remember that when a right tri-
r
angle has a hypotenuse of length r and an acute angle θ, as in the
r sin θ
picture on the right, the adjacent side will have length r cos θ and the
θ
opposite side will have length r sin θ. You can think of those lengths
r cos θ
as the horizontal and vertical “components” of the hypotenuse.
Notice in the above right triangle that we were given two pieces of information: one of
the acute angles and the length of the hypotenuse. From this we determined the lengths of
the other two sides, and the other acute angle is just the complement of the known acute
angle. In general, a triangle has six parts: three sides and three angles. Solving a triangle
means finding the unknown parts based on the known parts. In the case of a right triangle,
one part is always known: one of the angles is 90◦.
Example 1.19
Solve the right triangle in Figure 1.3.3 using the given
B
information:
c
a
(a) c = 10, A = 22◦
Solution: The unknown parts are a, b, and B. Solving yields:
A
b
C
a = c sin A = 10 sin 22◦ = 3.75
Figure 1.3.3
b = c cos A = 10 cos 22◦ = 9.27
B = 90◦ − A = 90◦ − 22◦ = 68◦
(b) b = 8, A = 40◦
Solution: The unknown parts are a, c, and B. Solving yields:
a = tan A ⇒ a = b tan A = 8 tan 40◦ = 6.71
b
b
b
8
= cos A
⇒
c =
=
= 10.44
c
cos A
cos 40◦
B = 90◦ − A = 90◦ − 40◦ = 50◦
(c) a = 3, b = 4
Solution: The unknown parts are c, A, and B. By the Pythagorean Theorem,
c =
a 2 + b 2 =
32 + 42 =
25 = 5 .
✄
Now, tan A = a
tan−1 on your
b = 3
4 = 0.75. So how do we find A? There should be a key labeled ✂
✁
calculator, which works like this: give it a number x and it will tell you the angle θ such that
tan θ = x. In our case, we want the angle A such that tan A = 0.75:
✄
Enter: 0.75
Press: tan−1
Answer: 36.86989765
✂
✁
This tells us that A = 36.87◦, approximately. Thus B = 90◦ − A = 90◦ − 36.87◦ = 53.13◦.
✄
✄
Note: The sin−1 and cos−1 keys work similarly for sine and cosine, respectively. These keys use
✂
✁
✂
✁
the inverse trigonometric functions, which we will discuss in Chapter 5.
20
Chapter 1 • Right Triangle Trigonometry
§1.3
Exercises
1. From a position 150 ft above the ground, an observer in a build-
12◦
ing measures angles of depression of 12◦ and 34◦ to the top and
34◦
bottom, respectively, of a smaller building, as in the picture on
150
the right. Use this to find the height h of the smaller building.
h
2. Generalize Example 1.12: A person standing a ft from the base of
a mountain measures an angle of elevation α from the ground to
the top of the mountain. The person then walks b ft straight back
h
and measures an angle of elevation β to the top of the mountain,
β
α
as in the picture on the right. Assuming the ground is level, find
b
a
a formula for the height h of the mountain in terms of a, b, α,
and β.
3. As the angle of elevation from the top of a tower to the sun decreases from 64◦ to 49◦ during the
day, the length of the shadow of the tower increases by 92 ft along the ground. Assuming the
ground is level, find the height of the tower.
4. Two banks of a river are parallel, and the distance between two
A
500
B
points A and B along one bank is 500 ft. For a point C on the
56◦
41◦
opposite bank, ∠ BAC = 56◦ and ∠ ABC = 41◦, as in the picture
w
on the right. What is the width w of the river?
( Hint: Divide AB into two pieces. )
C
5. A tower on one side of a river is directly east and north of
N
E
points A and B, respectively, on the other side of the river.
h
The top of the tower has angles of elevation α and β from A
and B, respectively, as in the picture on the right. Let d be
the distance between A and B. Assuming that both sides of
β
the river are at the same elevation, show that the height h of
α
the tower is
d
h =
.
(cot α)2 + (cot β)2
B
A
d
6. The equatorial parallax of the moon has been observed to be approximately 57′. Taking the radius
of the earth to be 3956.6 miles, estimate the distance from the center of the earth to the moon.
( Hint: See Example 1.15. )
7. An observer on earth measures an angle of 31′ 7′′ from one visible edge of the moon to the other
(opposite) edge. Use this to estimate the radius of the moon. ( Hint: Use Exercise 6 and see Example
1.16. )
Applications and Solving Right Triangles • Section 1.3
21
8. A ball bearing sits between two metal grooves, with the top groove having an
angle of 120◦ and the bottom groove having an angle of 90◦, as in the picture
1 ′′
120◦
2
on the right. What must the diameter of the ball bearing be for the distance
90◦
between the vertexes of the grooves to be half an inch? You may assume that
the top vertex is directly above the bottom vertex.
1.7
9. The machine tool diagram on the right shows a symmetric worm
thread, in which a circular roller of diameter 1.5 inches sits.
d
Find the amount d that the top of the roller rises above the
top of the thread, given the information in the diagram. ( Hint:
Extend the slanted sides of the thread until they meet at a point. )
1.5
10. Repeat Exercise 9 using 1.8 inches as the distance across the
top of the worm thread.
11. In Exercise 9, what would the distance across the top of the
30◦
worm thread have to be to make d equal to 0 inches?
12. For 0◦ < θ < 90◦ in the slider-crank mechanism in Example 1.18, show that
b 2 − r 2 (sin θ)2
c =
and
a = r sin θ +
b 2 − r 2 (sin θ)2 tan θ .
cos θ
( Hint: In Figure 1.3.2 draw line segments from B perpendicular to O A and AC. )
′′
13. The machine tool diagram on the right shows a symmetric die punch.
1 12
In this view, the rounded tip is part of a circle of radius r, and the slanted
54◦
r
sides are tangent to that circle and form an angle of 54◦. The top and
′′
2 18
bottom sides of the die punch are horizontal. Use the information in the
diagram to find the radius r.
B
14. In the figure on the right, ∠ BAC = θ and BC = a. Use this to find
a
AB, AC, AD, DC, CE, and DE in terms of θ and a.
( Hint: What is the angle ∠ ACD ? )
θ
A
C
D
E
For Exercises 15-23, solve the right triangle in Figure 1.3.4 using the
B
given information.
c
a
15. a = 5, b = 12
16. c = 6, B = 35◦
17. b = 2, A = 8◦
A
b
C
18. a = 2, c = 7
19. a = 3, A = 26◦
20. b = 1, c = 2
Figure 1.3.4
21. b = 3, B = 26◦
22. a = 2, B = 8◦
23. c = 2, A = 45◦
22
Chapter 1 • Right Triangle Trigonometry
§1.3
24. In Example 1.10 in Section 1.2, we found the exact values of all six trigonometric functions of
75◦. For example, we showed that cot 75◦ = 6− 2 . So since tan 15◦ = cot 75◦ by the Cofunction
6+ 2
Theorem, this means that tan 15◦ = 6− 2 . We will now describe another method for finding the
6+ 2
exact values of the trigonometric functions of 15◦. In fact, it can be used to find the exact values
for the trigonometric functions of θ when those for θ are known, for any 0◦
2
< θ < 90◦. The method
is illustrated in Figure 1.3.5 and is described below.
P
60◦
30◦
15◦
7.5◦
C
B
A
O
Q
1
Figure 1.3.5
Draw a semicircle of radius 1 centered at a point O on a horizontal line. Let P be the point on the
semicircle such that OP makes an angle of 60◦ with the horizontal line, as in Figure 1.3.5. Draw
a line straight down from P to the horizontal line at the point Q. Now create a second semicircle
as follows: Let A be the left endpoint of the first semicircle, then draw a new semicircle centered
at A with radius equal to AP. Then create a third semicircle in the same way: Let B be the left
endpoint of the second semicircle, then draw a new semicircle centered at B with radius equal to
BP.
This procedure can be continued indefinitely to create more semicircles. In general, it can be
shown that the line segment from the center of the new semicircle to P makes an angle with the
horizontal line equal to half the angle from the previous semicircle’s center to P.
(a) Explain why ∠ P AQ = 30◦. ( Hint: What is the supplement of 60◦ ? )
(b) Explain why ∠ PBQ = 15◦ and ∠ PCQ = 7.5◦.
(c) Use Figure 1.3.5 to find the exact values of sin 15◦, cos 15◦, and tan 15◦. ( Hint: To start, you
will need to use ∠ POQ = 60◦ and OP = 1 to find the exact lengths of PQ and OQ. )
(d) Use Figure 1.3.5 to calculate the exact value of tan 7.5◦.
(e) Use the same method but with an initial angle of ∠ POQ = 45◦ to find the exact values of
sin 22.5◦, cos 22.5◦, and tan 22.5◦.
Applications and Solving Right Triangles • Section 1.3
23
25. A manufacturer needs to place ten identical ball bearings against
r
the inner side of a circular container such that each ball bear-
ing touches two other ball bearings, as in the picture on the
right. The (inner) radius of the container is 4 cm.
4
(a) Find the common radius r of the ball bearings.
(b) The manufacturer needs to place a circular ring
inside the container. What is the largest possible
(outer) radius of the ring such that it is not on top
of the ball bearings and its base is level with the
base of the container?
26. A circle of radius 1 is inscribed inside a polygon with eight sides
of equal length, called a regular octagon. That is, each of the eight
sides is tangent to the circle, as in the picture on the right.
1
(a) Calculate the area of the octagon.
(b) If you were to increase the number of sides of the
polygon, would the area inside it increase or decrease?
What number would the area approach, if any? Explain.
(c) Inscribe a regular octagon inside the same circle. That is,
draw a regular octagon such that each of its eight vertexes
touches the circle. Calculate the area of this octagon.
27. The picture on the right shows a cube whose sides are of length
A
a > 0.
(a) Find the length of the diagona