Sine and Cosine Laws

1. Sine and Cosine Laws

2. Sine and Cosine Laws
Find x.
2
x
6
x – 4
x
6
4
x + 4
x
6
x + 4
x + 2
x8
10
B
9
A
4
Given the right triangle below
Find sin(A), cos(A), tan(A),
sin(B), cos(B) and tan(B).
A
64o
x
2x – 1
Find x.
π/3
1
x
π/6
40o
70o
15
x
Review problems

3. Sine and Cosine Laws
Every triangle can be split into
two right triangles,
joined at a common leg h.
A B
b a

4. Sine and Cosine Laws
Every triangle can be split into
two right triangles,
joined at a common leg h.
A
h
B
b a
h
B
a
A
hb

5. Sine and Cosine Laws
Every triangle can be split into
two right triangles,
joined at a common leg h.
From this split, we obtain
the Sine Law and the Cosine Law.
A
h
B
b a
h
B
a
A
hb

6. Sine and Cosine Laws
Every triangle can be split into
two right triangles,
joined at a common leg h.
From this split, we obtain
the Sine Law and the Cosine Law.
A
h
B
b a
The Sine Law h
B
a
A
hb

7. Sine and Cosine Laws
Every triangle can be split into
two right triangles,
joined at a common leg h.
From this split, we obtain
the Sine Law and the Cosine Law.
A
h
B
b a
We’ve h/b = sin(A) or h = b*sin(A),
The Sine Law
A
hb

8. Sine and Cosine Laws
Every triangle can be split into
two right triangles,
joined at a common leg h.
From this split, we obtain
the Sine Law and the Cosine Law.
A
h
B
b a
We’ve h/b = sin(A) or h = b*sin(A),
since h/a = sin(B) so h = a*sin(B)
so b*sin(A) = a*sin(B)
The Sine Law h
B
a
A
hb

9. Sine and Cosine Laws
Every triangle can be split into
two right triangles,
joined at a common leg h.
From this split, we obtain
the Sine Law and the Cosine Law.
A
h
B
b a
We’ve h/b = sin(A) or h = b*sin(A),
since h/a = sin(B) so h = a*sin(B)
so b*sin(A) = a*sin(B) or that
b
sin(B)
a
sin(A) =
The Sine Law h
B
a
A
hb

10. Sine and Cosine Laws
Every triangle can be split into
two right triangles,
joined at a common leg h.
From this split, we obtain
the Sine Law and the Cosine Law.
A
h
B
b a
We’ve h/b = sin(A) or h = b*sin(A),
since h/a = sin(B) so h = a*sin(B)
so b*sin(A) = a*sin(B) or that
b
sin(B)
a
sin(A) =
The Sine Law
The Sine Law: For any triangle,
sin(A) sin(B) sin(C)
= =
a b c
h
B
a
A
hb

11. Sine and Cosine Laws
Every triangle can be split into
two right triangles,
joined at a common leg h.
From this split, we obtain
the Sine Law and the Cosine Law.
A
h
B
b a
The Sine Law gives the precise meaning for
“the larger angle is opposite to the longer side.”
We’ve h/b = sin(A) or h = b*sin(A),
since h/a = sin(B) so h = a*sin(B)
so b*sin(A) = a*sin(B) or that
b
sin(B)
a
sin(A) =
h
B
a
A
hbThe Sine Law
The Sine Law: For any triangle,
sin(A) sin(B) sin(C)
= =
a b c

12. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
Sine and Cosine Laws

13. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
Sine and Cosine Laws
b =16

14. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
29A
Sine and Cosine Laws
b =16

15. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
29 74 CA
Sine and Cosine Laws
b =16

16. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
29 74A C
Sine and Cosine Laws
b =16

17. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
29 74A C
c
a
B
To solve the triangle:
Sine and Cosine Laws
b =16

18. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
<B = 180 – 29 – 74 = 77
29 74A
c
a
B
To solve the triangle:
Sine and Cosine Laws
b =16
C

19. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
sin(29) sin(77)
a 16
<B = 180 – 29 – 74 = 77
=
29 74A C
c
a
B
To solve the triangle:
Sine and Cosine Laws
b =16

20. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
sin(29) sin(77)
so a =
a 16
<B = 180 – 29 – 74 = 77
=
sin(77)
16sin(29)
29 74A C
c
a
B
To solve the triangle:
Sine and Cosine Laws
b =16

21. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
sin(29) sin(77)
so a =
a 16
<B = 180 – 29 – 74 = 77
=
sin(77)
16sin(29)
 7.96
29 74A C
c
a
B
To solve the triangle:
Sine and Cosine Laws
b =16

22. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
sin(29) sin(77)
so a =
a 16
<B = 180 – 29 – 74 = 77
=
sin(77)
16sin(29)
 7.96
sin(74) sin(77)
c 16
=
29 74A C
c
a
B
To solve the triangle:
Sine and Cosine Laws
b =16

23. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
sin(29) sin(77)
so a =
a 16
<B = 180 – 29 – 74 = 77
=
sin(77)
16sin(29)
 7.96
sin(74) sin(77)
c 16
=
so c =
sin(77)
16sin(74)
29 74A C
c
a
B
To solve the triangle:
Sine and Cosine Laws
b =16

24. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
sin(29) sin(77)
so a =
a 16
<B = 180 – 29 – 74 = 77
=
sin(77)
16sin(29)
 7.96
sin(74) sin(77)
c 16
=
so c =
sin(77)
16sin(74)
 15.8
29 74A C
c
a
B
To solve the triangle:
Sine and Cosine Laws
b =16

25. Example A. Solve for
the triangle given that
<A=29o, b=16, <C=74o.
Draw and label.
sin(29) sin(77)
so a =
a 16
<B = 180 – 29 – 74 = 77
=
sin(77)
16sin(29)
 7.96
sin(74) sin(77)
c 16
=
so c =
sin(77)
16sin(74)
 15.8
b =16
29 74A C
c
a
B
To solve the triangle:
Sine and Cosine Laws
Next we derive the Cosine Law
by splitting a triangle into two joined right triangles
and use the Pythagorean Theorem on them.

26. The Cosine Law
Sine and Cosine Laws

27. The Cosine Law
Sine and Cosine Laws
The cosine law gives the "adjustments“ to the
Pythagorean Theorem
for non right–triangles.

28. The Cosine Law
Sine and Cosine Laws
C
h
B
b
c
h
B
c
C
hb
a
x y
Divide a triangle into
two right triangles as shown
with let x and y be their legs
The cosine law gives the "adjustments“ to the
Pythagorean Theorem
for non right–triangles.

29. The Cosine Law
Sine and Cosine Laws
C
h
B
b
c
h
B
c
C
hb
a
x y+ = a
Divide a triangle into
two right triangles as shown
with let x and y be their legs
so that x + y = a or that y = a – x.
The cosine law gives the "adjustments“ to the
Pythagorean Theorem
for non right–triangles.

30. The Cosine Law
Sine and Cosine Laws
C
h
B
b
c
h
B
c
C
hb
a
x y+ = a
Divide a triangle into
two right triangles as shown
with let x and y be their legs
so that x + y = a or that y = a – x.
Applying the Pythagorean theorem
to both triangles and equating h2,
we have c2 – y2 = b2 – x2
The cosine law gives the "adjustments“ to the
Pythagorean Theorem
for non right–triangles.

31. The Cosine Law
Sine and Cosine Laws
C
h
B
b
c
h
B
c
C
hb
a
x y+ = a
Divide a triangle into
two right triangles as shown
with let x and y be their legs
so that x + y = a or that y = a – x.
Applying the Pythagorean theorem
to both triangles and equating h2,
we have c2 – y2 = b2 – x2 hence
c2 = b2 – x2 + y2
The cosine law gives the "adjustments“ to the
Pythagorean Theorem
for non right–triangles.

32. The Cosine Law
Sine and Cosine Laws
C
h
B
b
c
h
B
c
C
hb
a
x y+ = a
Divide a triangle into
two right triangles as shown
with let x and y be their legs
so that x + y = a or that y = a – x.
Applying the Pythagorean theorem
to both triangles and equating h2,
we have c2 – y2 = b2 – x2 hence
c2 = b2 – x2 + y2 setting y as (a – x)
c2 = b2 – x2 + (a – x)2 and expanding yield
The cosine law gives the "adjustments“ to the
Pythagorean Theorem
for non right–triangles.
c2 = b2 + a2 – 2*a*x

33. The Cosine Law
Sine and Cosine Laws
C
h
B
b
c
h
B
c
C
hb
a
x y+ = a
Divide a triangle into
two right triangles as shown
with let x and y be their legs
so that x + y = a or that y = a – x.
Applying the Pythagorean theorem
to both triangles and equating h2,
we have c2 – y2 = b2 – x2 hence
c2 = b2 – x2 + y2 setting y as (a – x)
c2 = b2 – x2 + (a – x)2 and expanding yield
The cosine law gives the "adjustments“ to the
Pythagorean Theorem
for non right–triangles.
c2 = b2 + a2 – 2*a*x or that c2 = b2 + a2 – 2*a*b*cos(C)
= x
b
–cos(C)

34. For any triangle,
c2 = a2 + b2 – 2*a*b*cos(C)
The Cosine Law:
Here are the cosine laws for solving sides a, b, or c.
Sine and Cosine Laws
b2 = a2 + c2 – 2*a*c*cos(B)
a2 = b2 + c2 – 2*b*c*cos(A)

35. For any triangle,
c2 = a2 + b2 – 2*a*b*cos(C)
The Cosine Law:
Here are the cosine laws for solving sides a, b, or c.
Sine and Cosine Laws
b2 = a2 + c2 – 2*a*c*cos(B)
a2 = b2 + c2 – 2*b*c*cos(A)
For solving angles A, B or C
we use the formulas:
a2 + b2 – c2
cos(C) =
2ab
a2 + c2 – b2
cos(B) =
2ac
b2 + c2 – a2
cos(A) =
2bc

36. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
Sine and Cosine Laws

37. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
Sine and Cosine Laws
a=29

38. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
Sine and Cosine Laws
a=29
16o
B
c=74

39. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
Sine and Cosine Laws
a=29
16o
B
c=74
b

40. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
Using the Cosine Law for side b:
b2 = 292 + 742 – [ 2(29)(74)cos(16) ]
Sine and Cosine Laws
a=29
16o
B
c=74
b

41. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
b = 292 + 742 – [ 2(29)(74)cos(16) ]
Using the Cosine Law for side b:
b2 = 292 + 742 – [ 2(29)(74)cos(16) ]
Sine and Cosine Laws
a=29
16o
B
c=74
b

42. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
b = 292 + 742 – [ 2(29)(74)cos(16) ]  46.8
Using the Cosine Law for side b:
b2 = 292 + 742 – [ 2(29)(74)cos(16) ]
Sine and Cosine Laws
a=29
16o
B
c=74
b

43. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
b = 292 + 742 – [ 2(29)(74)cos(16) ]  46.8
Using the Cosine Law for side b:
b2 = 292 + 742 – [ 2(29)(74)cos(16) ]
Sine and Cosine Laws
However if we are given that
<B=16o, c=74 and that b  46.8
there are two possibilities for
side a. We see this if we
construct the triangles.
a=29
16o
B
c=74
b

44. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
b = 292 + 742 – [ 2(29)(74)cos(16) ]  46.8
Using the Cosine Law for side b:
b2 = 292 + 742 – [ 2(29)(74)cos(16) ]
Sine and Cosine Laws
However if we are given that
<B=16o, c=74 and that b  46.8
there are two possibilities for
side a. We see this if we
construct the triangles.
c=74
16oB
a=29
16o
B
c=74
b

45. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
b = 292 + 742 – [ 2(29)(74)cos(16) ]  46.8
Using the Cosine Law for side b:
b2 = 292 + 742 – [ 2(29)(74)cos(16) ]
Sine and Cosine Laws
However if we are given that
<B=16o, c=74 and that b  46.8
there are two possibilities for
side a. We see this if we
construct the triangles.
c=74
16oB
46.846.8
a=29
16o
B
c=74
b

46. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
b = 292 + 742 – [ 2(29)(74)cos(16) ]  46.8
Using the Cosine Law for side b:
b2 = 292 + 742 – [ 2(29)(74)cos(16) ]
Sine and Cosine Laws
However if we are given that
<B=16o, c=74 and that b  46.8
there are two possibilities for
side a. We see this if we
construct the triangles.
c=74
16oB
46.846.8
c=74
16
46.8c=74
16
46.8
C C
a=29
16o
B
c=74
b

47. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
b = 292 + 742 – [ 2(29)(74)cos(16) ]  46.8
Using the Cosine Law for side b:
b2 = 292 + 742 – [ 2(29)(74)cos(16) ]
Sine and Cosine Laws
However if we are given that
<B=16o, c=74 and that b  46.8
there are two possibilities for
side a. We see this if we
construct the triangles.
c=74
16oB
46.846.8
c=74
16
46.8c=74
16
46.8
Using sine-law to solve for C, we get sin(C)  0.435.
C C
a=29
16o
B
c=74
b

48. Example B. Given that
a=29, <B=16o, c=74.
draw and find side b.
b = 292 + 742 – [ 2(29)(74)cos(16) ]  46.8
Using the Cosine Law for side b:
b2 = 292 + 742 – [ 2(29)(74)cos(16) ]
Sine and Cosine Laws
However if we are given that
<B=16o, c=74 and that b  46.8
there are two possibilities for
side a. We see this if we
construct the triangles.
c=74
16oB
46.846.8
c=74
16
46.8c=74
16
46.8
Using sine-law to solve for C, we get sin(C)  0.435.
Given that C + 16 <180, the solutions for C are
C  25.8o or 180 – 25.8o 154.2o as shown.
C C
a=29
16o
B
c=74
b

49. Sine and Cosine Laws
Ex. Find all the missing sides and angles.
1. 3.2.
4. 6.5.
7. 9.8.
10. 12.11.
12
13
6
9
A
76o
C
D
x
B

50. Sine and Cosine Laws
15. Two airplanes leave Berlin, one heading straight for London
and the other straight for Paris. The angle formed is 18 degrees.
Use the Law of Cosines to estimate the distance from London to
Paris.
13. A baseball infield is determined by a square with sides 90 ft
long. In the diagram, home plate is H and first base is F.
Suppose the first baseman ran in a straight line from F to catch
a pop-up at B, 120 ft from home plate. If the measure of
FHB is 10°, how far did the first baseman run?
14. Some students in Geometry are assigned the task of measuring
the distance between two trees separated by a swamp. The students
determine that the angle formed by tree A, a dry point C, and tree B
is 27°. They also know that mABC is 85°. If AC is 150 ft, how
far apart are the trees?
16. Two lookout towers, L and M, are 50 kilometers apart. The ranger in Tower
L sees a fire at point C such that mCLM = 40°. The ranger in Tower M sees
the same fire such that mCML = 65°. How far is the fire from Tower L?

51. Sine and Cosine Laws
Find the indicated angles and sides in the following figure.
4
17.
3
4
5A
21.
3
B
C
B
4
5
A
3
B22. Find the angles A, B
and C between the
diagonal and the three
edges as shown.
C
6
8
A B
C
18. 1
5A
19.
20.
B
Ax
x
y
x
y
z
x
y C
C

52. Sine and Cosine Laws
Answers
A. 1. r ≈ 37.4, A ≈ 35.4o, T ≈ 24.6o
3. a ≈ 25.8, d ≈ 23.7, A = 70o
5. p ≈ 118, t ≈ 181, P = 36o
7. H ≈ 23.3o, I ≈ 97.4o, G ≈ 59.3 o
9. r ≈ 8.53, a ≈ 19.3, A = 107o
11. c ≈ 17.6, a ≈ 19.1, C = 67o
13. ≈ 35 ft 15. ≈ 290 mi
17. C ≈ 73.7o, A = B ≈ 97.4o
19. A ≈ 23.3o, B ≈ 57.7o
21. A ≈ 72o, B ≈ 48o, C ≈ 60o,