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section 6.2 The Law of Sines and the Law of Cosines 475
of a triangle, then finding the third angle is easy because the sum of the
angles of a triangle equals π radians (or equivalently 180◦ if we are working
in degrees).
When to use which law
Use the law of cosines if you know
• the lengths of all three sides of a triangle;
• the lengths of two sides of a triangle and the angle between them.
Use the law of sines if you know
• two angles of a triangle and the length of one side;
• the length of two sides of a triangle and an angle other than the
angle between those two sides.
exercises
In Exercises 1–16 use the following figure (which
is not drawn to scale). When an exercise requests
that you evaluate an angle, give answers in both
radians and degrees.
5.
Suppose a = 3, b = 5, and c = 6. Evaluate:
(a) A
6.
7.
c
8.
A
9.
Suppose a = 6, B = 25◦ , and C = 40◦ .
Evaluate:
(a) A
2.
(b) b
Suppose a = 6, A =
radians. Evaluate:
(a) C
4.
(c) c
11.
(b) b
Suppose a = 4, B =
radians. Evaluate:
(a) A
π
7
2π
11
(b) b
12.
radians, and B =
4π
7
radians, and C =
(c) c
3π
11
(c) C
(b) A
(c) B
(b) A
(c) B
(b) A
(c) B
Suppose a = 4, b = 5, and C = 2 radians.
Evaluate:
(a) c
13.
(c) c
(b) B
Suppose a = 3, b = 4, and C = 1 radian.
Evaluate:
(a) c
(c) c
(c) C
Suppose a = 5, b = 7, and C = 23◦ . Evaluate:
(a) c
Suppose a = 7, B = 50◦ , and C = 35◦ .
Evaluate:
(a) A
3.
(b) b
10.
(b) B
Suppose a = 2, b = 3, and C = 37◦ . Evaluate:
(a) c
1.
(c) C
Suppose a = 6, b = 7, and c = 8. Evaluate:
(a) A
C
b
(b) B
Suppose a = 5, b = 6, and c = 9. Evaluate:
(a) A
a
(c) C
Suppose a = 4, b = 6, and c = 7. Evaluate:
(a) A
B
(b) B
(b) A
(c) B
Suppose a = 4, b = 3, and B = 30◦ . Evaluate:
(a) A (assume that A < 90◦ )
(b) C
(c) c
476
14.
chapter 6 Applications of Trigonometry
Suppose a = 14, b = 13, and B = 60◦ .
Evaluate:
(a) A (assume that A < 90◦ )
(b) C
16.
Suppose a = 14, b = 13, and B = 60◦ .
Evaluate:
(a) A (assume that A > 90◦ )
(c) c
15.
[Exercises 15 and 16 should be compared with
Exercises 13 and 14.]
Suppose a = 4, b = 3, and B = 30◦ . Evaluate:
◦
(a) A (assume that A > 90 )
(b) C
(c) c
(b) C
(c) c
problems
17. Write the law of sines in the special case of a
right triangle.
18. Show how the previous problem gives the familiar characterization of the sine of an angle
in a right triangle as the length of the opposite
side divided by the length of the hypotenuse.
19. Show how Problem 17 gives the familiar characterization of the tangent of an angle in a right
triangle as the length of the opposite side divided by the length of the adjacent side.
20. Suppose a triangle has sides of length a, b, and
c satisfying the equation
a2 + b 2 = c 2 .
Show that this triangle is a right triangle.
21. Show that in a triangle whose sides have
lengths a, b, and c, the angle between the sides
of length a and b is an acute angle if and only
if
a2 + b 2 > c 2 .
22. Show that
r
p= √ √
2 1 − cos θ
in an isosceles triangle that has two sides of
length p, an angle of θ between these two
sides, and a third side of length r .
23. Use the law of cosines to show that if a, b, and
c are the lengths of the three sides of a triangle, then
c 2 > a2 + b2 − 2ab.
24. Use the previous problem to show that in every triangle, the sum of the lengths of any two
sides is greater than the length of the third
side.
25. Suppose you need to walk from a point P to a
point Q. You can either walk in a line from P to
Q, or you can walk in a line from P to another
point R and then walk in a line from R to Q.
Use the previous problem to determine which
of these two paths is shorter.
26. Suppose you are asked to find the angle C
formed by the sides of length 2 and 3 in a triangle whose sides have length 2, 3, and 7.
(a) Show that in this situation the law of
cosines leads to the equation cos C = −3.
(b) There is no angle whose cosine equals −3.
Thus part (a) seems to give a counterexample to the law of cosines. Explain what
is happening here.
27. The law of cosines is stated in this section using the angle C. Using the labels of the triangle
just before Exercise 1, write two versions of the
law of cosines, one involving the angle A and
one involving the angle B.
28. Use one of the examples from this section to
show that
cos−1
1
5
+ cos−1
5
7
+ cos−1
19
35
= π.
29. Show that
a(sin B − sin C) + b(sin C − sin A) + c(sin A − sin B)
=0
in a triangle with sides whose lengths are a, b,
and c, with corresponding angles A, B, and C
opposite those sides.
section 6.2 The Law of Sines and the Law of Cosines 477
30. Show that
31. Show that
c = b cos A + a cos B
a2 + b2 + c 2 = 2(bc cos A + ac cos B + ab cos C)
in a triangle with sides whose lengths are a, b,
and c, with corresponding angles A, B, and C
opposite those sides.
[Hint: Add together the equations a2 =
b2 + c 2 − 2bc cos A and b2 = a2 + c 2 − 2ac cos B.]
in a triangle with sides whose lengths are a, b,
and c, with corresponding angles A, B, and C
opposite those sides.
worked-out solutions to Odd-numbered Exercises
In Exercises 1–16 use the following figure (which
is not drawn to scale). When an exercise requests
that you evaluate an angle, give answers in both
radians and degrees.
(c) Use the law of sines in the form
sin A
sin C
=
,
a
c
which in this case becomes the equation
sin 115◦
sin 40◦
=
.
6
c
B
c
Solve the equation above for c, getting
a
c=
A
C
b
1.
3.
Suppose a = 6, B = 25◦ , and C = 40◦ .
Evaluate:
(a) A
(b) b
(c) c
solution
◦
(a) The angles in a triangle add up to 180 . Thus
A + B + C = 180◦ . Solving for A, we have
A = 180◦ − B − C = 180◦ − 25◦ − 40◦ = 115◦ .
π
180◦
Multiplying by
A = 115◦ =
23π
36
to convert to radians gives
radians ≈ 2.007 radians.
(b) Use the law of sines in the form
Suppose a = 6, A =
radians. Evaluate:
(a) C
sin 25◦
sin 115◦
=
.
6
b
Solve the equation above for b, getting
6 sin 25◦
≈ 2.80.
b=
sin 115◦
π
7
radians, and B =
(b) b
4π
7
(c) c
solution
(a) The angles in a triangle add up to π radians.
Thus A + B + C = π . Solving for C, we have
C =π −A−B =π −
Multiplying by
180◦
π
C=
π
7
−
4π
7
=
2π
7
.
to convert to radians gives
2π
7
radians =
360 ◦
.
7
Using a calculator to obtain decimal approximations, we have
C ≈ 0.8976 radians ≈ 51.429◦ .
(b) Use the law of sines in the form
sin A
sin B
=
,
a
b
sin B
sin A
=
,
a
b
which in this case becomes the equation
6 sin 40◦
≈ 4.26.
sin 115◦
which in this case becomes the equation
sin
π
7
6
=
sin
4π
7
b
.
Solve the equation above for b, getting
b=
6 sin
sin
4π
7
π
7
≈ 13.48.