Download 10.6A The Law of Cosines

10.6A The Law of Cosines
Objectives:
G.SRT.10: Prove the Laws of Sines and Cosines and use them to solve problems.
G.SRT.11: Understand and apply the Law of Sines and the Law of Cosines to find unknown
measurements in right and non-right triangles.
For the Board: You will be able to use the Law of Cosines to find the side lengths and angle measures
of a triangle and to use Heron’s Formula to find the area of a triangle.
Anticipatory Set:
The Law of Sines is used to solve triangles which have a AAS, ASA, or SSA pattern.
The Law of Cosines is used to solve triangle which have a SSS or SAS pattern.
Instruction:
Law of Cosines
For ΔABC, the Law of Cosines states that
a2 = b2 + c2 – 2bc cos A
b2 = a2 + c2 – 2ac cos B
c2 = a2 + b2 – 2ab cos C
B
c
A
a
C
b
Open the book to page 730 – 731 and read example 1.
Examples: Use the given measurements to solve ΔABC. Round to the nearest tenth.
A. a = 8, b = 5, m<C = 32.2°
1. Draw an accurate sketch. Remember that short sides
are across from small angles and large sides are
across from large angles.
2. When given an angle measure, use the
version of the Law of Cosines for that angle.
c2 = a2 + b2 – 2ab cos C
c2 = 82 + 52 – 2(8)(5) cos 32.2° = 21.3
c = 4.6
C
B
8
c
32.2°
5
A
3. Never use the Law of Sines to find an obtuse angle.
Use the Law of Sines to find m<B not m<A because m<A appears to be obtuse.
sin A sin B sin C
sin A sin B sin 32.2




a
b
c
8
5
4.6
4.6 sin B = 5 sin 32.2
sin B = 0.579213344
sin-1(0.579213344) = 35.4°
4. Find m<A using the 3 angle of a triangle add to 180°.
m<A = 180° – 32.2° – 35.4° = 112.4°
B. a = 8, b = 9, c = 7
B
1. Draw an accurate sketch.
Recall: Large angles are across from large sides.
If the triangle has an obtuse angle, it will
be across from b = 9.
7
A
8
9
2. When given a choice use the Law of Cosines to find the angle which might be
obtuse, since the Law of Sines does not work for obtuse angles.
Find m<B.
b2 = a2 + c2 – 2ac cos B
92 = 82 + 72 – 2(8)(7) cos B
81 = 64 + 49 – 224 cos B
-32 = -112 cos B
cos B = 0.2857142863
m<B = Cos-1 0.2857142863 = 73.4°
3. Use the Law of Sines to find either m<A or m<C.
sin A sin B sin C


a
b
c
sin A sin 73.4 sin C


8
9
7
8 sin 73.4 = 9 sin A
sin A = 0.95819
m< A = Sin-1 0.851842288 = 58.4°
4. Find the remaining angle.
m<C = 180° – 73.4° – 58.4° = 48.2°
White Board Activity:
Practice: a. b = 23, c = 18, m<A = 173°
Use the version with m<A: a2 = b2 + c2 – 2bc cos A
a2 = 232 + 182 – 2(23)(18) cos 173° = 1674.828
a = 40.9
Use the Law of sines to find either m<B or m<C since both are acute.
sin A sin B sin C
sin 173 sin B sin C




a
b
c
40.9
23
18
23 sin 173 = 40.9 sin B
sin B = 0.0685329
-1
m<B = Sin 0.0685329 = 3.9°
Use the sum of the angles of a triangle is 180° to find the remaining angle.
m<C = 180° – 173° – 3.9° = 3.1°
C
b. a = 35, b = 42, c = 50.3
Since c is the largest side, if there is an obtuse angle in the triangle it will be <C.
Use the version with m<C.
c2 = a2 + c2 – 2ac cos C
50.32 = 352 + 422 – 2(35)(42) cos C
2530.09 = 1225 + 1764 – 2940 cos C
-458.91 = -2940 cos C
cos C = 0.156091837
m<C = Cos-1 0.156091837 = 81.0°
Use the Law of sines to find either <A or <B since they are both acute.
sin A sin B sin C
sin A sin B sin 81.0




a
b
c
35
42
50.3
35 sin 81.0 = 50.3 sin A
sin A = 0.687258289
m< B = Sin-1 0. 687258289 = 43.4°
Use the sum of the angles of a triangle is 180° to find the remaining angle.
m<B = 180° – 81.0° – 43.4 ° = 55.6°
Assessment:
Question student pairs.
Independent Practice:
Text: pgs. 734 – 735 prob. 1 – 6, 9 – 14, 17 – 22.
For a Grade:
Text: pg. 734 – 735 prob. 4, 10.