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Mr. Smith’s IM 3
IM 3 Assignment 6.2: Worked Out Examples
Law of Sines and Cosines.
Examples.
1. Side a = 3, side b = 5, and side c = 7. Find all missing angles. In this case,
3 sides are known so you should use the law of cosines.
c2 = a2 + b2 - 2ab Cos C
49 = 9 + 25 - 2(3)(5) Cos C
49 = 34 - 30 Cos C
15 = -30 Cos C
-1/2 = Cos C
Cos-1 (-1/2) = C
120 = C
C = 120
b2 = a2 + c2 – 2ac CosB
25 = 9 + 49 – 2(21) CosB
25 = 58 – (42) CosB
-33 = (-42) CosB
.7857 = CosB
Cos-1 (.7857) = B
38 = B
B = 38
A2 = b2 + c2 – 2bc CosA
9 = 25 + 49 – 2(5)(7) CosA
9 = 74 – 2(35) CosA
-65 = (-70) CosA
.92857 = CosA
Cos-1 (.92857) = A
Mr. Smith’s IM 3
22 = A
A = 22
2. Side a = 4, side b = 6, and angle A = 40. Find angle B. In this case, 2
sides and an angle opposite one of the sides are known, so you can use the
law of sines.
a = b
Sin A Sin B
4 = 6
Sin 40 Sin B
4 (Sin B) = 6 (Sin 40)
Sin B = 6 (Sin 40)
4
Sin B = 0.964
B = 75
3. If side a = 8, side b = 11, and side c = 14, what are the missing angles?
You again have three sides, so the law of cosines should be used.
b2 = a2 + c2 - 2ac CosB
121 = 64 + 196 – 2(112) CosB
121 = 260 – (224) CosB
-139 = -224 CosB
.6205 = CosB
Cos-1 (.6205) = B
Mr. Smith’s IM 3
52 = B
B = 52
a2 = b2 + c2 – 2bc CosA
64 = 121 + 196 – 2(154) CosA
64 = 317 – (308) CosA
-253 = (-308) CosA
.8214 = CosA
Cos-1 (.8214) = A
34.77 = A
A = 35
Now angle C can be found from subtracting 35 and 52 from 180, this
equals
180 - 35 – 52 = 93
(However you can check that this is right by again using the law of cosines).
c2 = a2 + b2 – 2ab CosC
196 = 64 + 121 – 2(88) CosC
196 = 185 – (176) CosC
11 = (- 176) CosC
-.0625 = CosC
Cos-1 (-.0625) = C
93 = C
C = 93
4. For this triangle, the sides a = 1.5, b = 1.7, and angle B = 46. Use the law
of Sines to find angle A.
Mr. Smith’s IM 3
a = b
Sin A Sin B
1.5 = 1.7
Sin A Sin46
1.7SinA = 1.5Sin46
SinA = 1.5Sin46
1.7
Sin A = 0.6347
A = Sin-1 0.6347
A = 39
5. You can also apply the law of sines and cosines to other shapes when
trying to find area, in such things as the example shown below. First you
must separate the shape into two triangles. Then you should find Angle C in
the other triangle because it it needed in the formula for area, which is Area
= ( ½ (ab) SinC)
First using the Law of Cosines, we can find the missing side.
c2 = a2 + b2 – 2ab CosC
c2 = 47 + 52 – 2(47)(52) Cos98
c2 = 2209 + 2704 – 2(2444) Cos98
c2 = 4913 – 4888 Cos98
c2 = 4913 + 680.28
c2 = 5593.28
c = 74.8
Now that you know the measure of side c, you can again use the Law of
Cosines to find the opposite angle being Angle C.
c2 = a2 + b2 – 2ab CosC
74.82 = 392 + 722 – 2(39)(72) CosC
5595 = 1521 + 5184 – 2(2808) CosC
5595 = 6705 – 5616 CosC
Mr. Smith’s IM 3
-1110 = -5616 CosC
0.1976 = CosC
Cos-1 0.1976 = C
C = 79
Another formula that can be used to find an angle using the Law of Cosines
is CosC = a2 + b2 – c2
2 (a)(b)
Now using the area formula we can figure out the area of the two triangles
and combine the areas.
Area(1) = ½ (ab) SinC
Area = ½ (47)(52) Sin98
Area = ½ 2444 Sin98
Area = ½ 2420
Area(1) = 12102
Area(2) = ½ (ab) SinC
Area = ½ (39)(72) Sin 79
Area = ½ (2808) Sin 79
Area = ½ 2756
Area(2) = 13782
Now by combining area(1) and area(2) you get the total area.
Area(1) + Area(2) = Area(T)
12102 + 13782 = 25882
The total area equals 2588 units squared.
6. Other examples to get you familiar with using both the Law of Sines as
well as the Law of Cosines are in such triangles where the three sides are
known, in this case side a = 5, side b = 6, and side c = 8, you should find all
angles. Knowing all sides means that you must use the Law of Cosines to
Mr. Smith’s IM 3
start.
a2 = b2 + c2 – 2 bc CosA
52 = 62 + 82 – 2 (8)(6) CosA
25 = 36 + 64 – 2 (48) Cos A
25 = 100 – 96 Cos A
-75 = -96 CosA
0.78125 = CosA
Cos-1 0.78125 = A
A = 39
You now know one of the angles of the triangle, so you can apply the Law
of Sines to find the other missing angles.
a = b
SinA SinB
5 = 6
Sin39 SinB
5SinB = 6Sin39
SinB = 6Sin39
5
SinB = 0.7552
B =Sin-1 0.7552
B = 49
And since you now know two of three possible angles, you know that a
triangle is made up to 180, therefore you can subtract the two known angles
from 180 to get the final missing angle.
180 – 49 – 39 = 92
C = 92
7. For this triangle you are given two sides and one angle between the two
known sides, from the known information you have, you now know you
should use the Law of Cosines to find the missing information which in this
case is side c, angle A, and angle B. Side a = 12, side b = 14, and angle C =
110.
Mr. Smith’s IM 3
c2 = a2 + b2 – 2ab CosC
c2 = 12 + 14 – 2 (12)(14) Cos110
c2 = 144 + 196 – 2 (168) Cos110
c2 = 340 – 336 Cos110
c2 = 340 + 115
c2 = 455
c = 21.3
Now, three sides and an agle opposite one of the sides are known, so you
could decide whether to use the Law of Sines or the Law of Cosines, but the
Law of Sines is easier to work with.
a = c
SinA SinC
12 = 21.33
SinA Sin110
21.33 SinA = 12Sin110
SinA = 12Sin110
21.33
SinA = 0.5287
A = Sin 0.5287
A = 32
Now Angle B can again be found from subtracting angle A and angle C
from 180.
180 – 110 – 32 = 38
B = 38
8. Just to become more familiar, we will show you one more example of
how to find the area of a non-triangular shape, being a quadrilateral in this
Mr. Smith’s IM 3
case.
Now that the two triangles have been formed we can proceed to use the Law
of Cosines to find the length of side c.
c2 = a2 + b2 – 2ab CosC
c2 = 102 + 182 – 2 (10)(18) Cos85
c2 = 100 + 324 – 2 (180) Cos85
c2 = 424 – 360 Cos85
c2 = 424 – 31.38
c2 = 392.62
c = 19.8
Now again you must find the measurement for the missing angle on the
other triangle in order to solve for the area. Again you can use the Law of
Cosines.
c2 = a2 + b2 – 2ab CosC
19.82 = 142 + 162 – 2(14)(16) CosC
392.04 = 196 + 256 – 2(224) CosC
392.04 = 452 - 448 CosC
-59.96 = -448 CosC
0.1338 = CosC
Cos-1 0.1338 = C
C = 82
Now that you know the angles opposite the side separating the two triangles,
you can figure out the area of the entire figure.
Area(1) = ½ (ab) SinC
Area = ½ (10)(18) Sin85
Area = ½ 180 Sin85
Area = ½ 179.32
Area(1) = 89.662
Area(2) = ½ (ab) SinC
Area = ½ (14)(16) Sin 82
Area = ½ (224) Sin 82
Area = ½ 221.82
Area(2) = 110.912
Mr. Smith’s IM 3
Area(1) + Area(2) = Area(T)
89.66 + 110.91 = 200.572
Total area equals 200.57 units squared.