9.6 Apply the Law of Cosines In which cases can the law of cosines be used to solve a triangle? What...
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9.6 Apply the Law of Cosines In which cases can the law of cosines be used to solve a triangle? What is Heron’s Area Formula? What is the semiperimeter?
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Transcript of 9.6 Apply the Law of Cosines In which cases can the law of cosines be used to solve a triangle? What...
9.6 Apply the Law of Cosines
In which cases can the law of cosines be used to solve a triangle?
What is Heron’s Area Formula?What is the semiperimeter?
Law of Cosines
Use the law of cosines to solve triangles when two sides and the included angle are known (SAS), or when all three side are known (SSS).
Solve ABC with a = 11, c = 14, and B = 34°.
SOLUTION
Use the law of cosines to find side length b.
b2 = a2 + c2 – 2ac cos B
b2 = 112 + 142 – 2(11)(14) cos 34°
b2 61.7
b2 61.7 7.85
Law of cosines
Substitute for a, c, and B.
Simplify.
Take positive square root.
Use the law of sines to find the measure of angle A.
sin Aa
sin Bb
=
sin A11
=sin 34°7.85
sin A =11 sin 34°
7.850.7836
A sin –1 0.7836 51.6°
Law of sines
Substitute for a, b, and B.
Multiply each side by 11 andSimplify.
Use inverse sine.
The third angle C of the triangle is C 180° – 34° – 51.6° = 94.4°.
In ABC, b 7.85, A 51.68, and C 94.48.ANSWER
Solve ABC with a = 12, b = 27, and c = 20.
SOLUTION
First find the angle opposite the longest side, AC . Use the law of cosines to solve for B.
b2 = a2 + c2 – 2ac cos B
272 = 122 + 202 – 2(12)(20) cos B
272 = 122 + 202
– 2(12)(20)= cos B
– 0.3854 cos BB cos –1 (– 0.3854) 112.7°
Law of cosines
Substitute.
Solve for cos B.
Simplify.
Use inverse cosine.
Now use the law of sines to find A.
sin Aa =
sin Bb
sin A12
sin 112.7°27
=
sin A =12 sin 112.7°
270.4100
A sin–1 0.4100 24.2°
Law of sines
Substitute for a, b, and B.
Multiply each side by 12 and simplify.
Use inverse sine.
The third angle C of the triangle is C 180° – 24.2° – 112.7° = 43.1°.
In ABC, A 24.2, B 112.7, and C 43.1.ANSWER
ScienceScientists can use a set of footprints to calculate an organism’s step angle, which is a measure of walking efficiency. The closer the step angle is to 180°, the more efficiently the organism walked.
The diagram at the right shows a set of footprints for a dinosaur. Find the step angle B.
SOLUTION b2 = a2 + c2 – 2ac cos B3162 = 1552 + 1972 – 2(155)(197) cos B
3162 = 1552 + 1972
– 2(155)(197)= cos B
– 0.6062 cos B
B cos –1 (– 0.6062) 127.3° Use inverse cosine.
Simplify.
Solve for cos B.
Substitute.
Law of cosines
The step angle B is about 127.3°.ANSWER
Find the area of ABC.
1. a = 8, c = 10, B = 48°
SOLUTION
Use the law of cosines to find side length b.
b2 = a2 + c2 – 2ac cos B
b2 = 82 + 102 – 2(8)(10) cos 48°
b2 57
b2 57 7.55
Law of cosines
Substitute for a, c, and B.
Simplify.
Take positive square root.
Use the law of sines to find the measure of angle A.
sin Aa
sin Bb
=
sin A 8
=sin 48°7.55
sin A =8 sin 48°
7.550.7874
A sin –1 0.7836 51.6°
Law of sines
Substitute for a, b, and B.
Multiply each side by 8 andsimplify.
Use inverse sine.
The third angle C of the triangle is C 180° – 48° – 51.6° = 80.4°.
In ABC, b 7.55, A 51.6°, and C 80.4°.ANSWER
162 = 142 + 92 – 2(14)(9) cos B
Find the area of ABC. 2. a = 14, b = 16, c = 9
SOLUTION First find the angle opposite the longest side, AC . Use the law of cosines to solve for B.
b2 = a2 + c2 – 2ac cos B
162 = 142 + 92
– 2(14)(9)= cos B
Law of cosines
Substitute.
Solve for cos B.
– 0.0834 cos BB cos –1 (– 0.0834) 85.7°
Simplify.
Use inverse cosine.
sin Aa
= sin Bb
sin A14
sin 85.2°16
=
sin A =14sin 85.2°
160.8719
Law of sines
Substitute for a, b, and B.
Multiply each side by 14 and simplify.
Use the law of sines to find the measure of angle A.
The third angle C of the triangle is C 180° – 85.2° – 60.7° = 34.1°.
A sin–1 0.8719 60.7° Use inverse sine.
In ABC, A 60.7°, B 85.2°, and C 34.1°.
ANSWER
Heron’s Area FormulaHeron (Hero) of Alexandria, the Greek mathematician (10 - 70 A.D.) is credited with using the law of cosines to find this formula for the area of a triangle.
Urban PlanningThe intersection of three streets forms a piece of land called a traffic triangle. Find the area of the traffic triangle shown.
SOLUTION STEP 1 Find the semiperimeter s.
s = (a + b + c )12
12= (170 + 240 + 350) = 380
STEP 2 Use Heron’s formula to find the area of ABC.
Area = s (s – a) (s – b) (s – c)
380 (380 – 170) (380 – 240) (380 – 350)= 18,300
The area of the traffic triangle is about 18,300 square yards.
Find the area of ABC. 4.
SOLUTION
STEP 1 Find the semiperimeter s.
s = (a + b + c )12
12= (5 + 8 + 11) = 12
STEP 2 Use Heron’s formula to find the area of ABC.
Area = s (s – a) (s – b) (s – c)
The area is about 18.3 square units.
12 (12 – 8) (12 – 11) (12 – 5)= 18.3
Find the area of ABC. 5.
SOLUTION
STEP 1 Find the semiperimeter s.
s = (a + b + c )12
12= (4 + 9+ 7) = 10
STEP 2 Use Heron’s formula to find the area of ABC.
Area = s (s – a) (s – b) (s – c)
The area is about 13.4 square units.
10 (10– 4) (10 – 9) (10 – 7)= 13.4
Find the area of ABC.6.
SOLUTION
STEP 1 Find the semiperimeter s.
s = (a + b + c )12
12= (15 + 23+ 127) = 25
STEP 2 Use Heron’s formula to find the area of ABC.
Area = s (s – a) (s – b) (s – c)
The area is about 80.6 square units.
25 (25– 15) (25 – 23) (25 – 12)= 80.6
9.6 Assignment
Page 596, 3-39 every 3rd problem No work is the same as a missing problem
