Transcript Document

Lesson 35 – The Cosine Law
Integrated Math 10 – Santowski
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(A) Cosine Law - Derivation
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Cosine function for triangle
ADB
cosA=c/x  x=ccosA
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Pythagorean theorem for
triangle ADB
x2+h2=c2  h2=c2−x2
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Pythagorean theorem for
triangle CDB
(b−x)2+h2=a2
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(A) Cosine Law - Derivation
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Pythagorean theorem for triangle CDB
(b−x)2+h2=a2
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Substitute h2 = c2 - x2
(b−x)2+(c2−x2)=a2
(b2−2bx+x2)+(c2−x2)=a2
b2−2bx+c2=a2
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Substitute x = c cos A
b2−2b(ccosA)+c2=a2
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Rearrange:
a2=b2+c2−2bccosA
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(C) Law of Cosines:
A
b
Have: two sides,
included angle
Solve for: missing side
2
c
=
2
a
+
C
2
b
c
a
B
– 2 a b cos C
(missing side)2 = (one side)2 + (other side)2 – 2 (one side)(other side) cos(included angle)
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(C) Law of Cosines:
A
Have: three sides
b
c
Solve for: missing angle
C
B
Side Opposite
Missing Angle
Missing Angle
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a
a2 + b2 – c2
cos C =
2ab
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(D) Cosine Law - Examples
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Solve this triangle
B
c=5.2
a=2.4
A
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b=3.5
Integrated Math 10 - Santowski
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(D) Cosine Law - Examples
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B
c=5.2
a=2.4
A
b=3.5
C
Start with the law of cosines because there are no angles given.
a2=b2+c2-2bc cosA. Substitute values. 2.42=3.52+5.22-2(3.5)(5.2) cosA,
5.76-12.25-27.04=-2(3.5)(5.2) cos A, 33.53=36.4cosA, 33.53/36.4=cos A, 0.921=cos A, A=67.07.
Now for B.
b2=a2+c2-2ac cosB, (3.5)2=(2.4)2+(5.2)2-2(2.4)(5.2) cosB, 12.25=5.76+27.04-24.96 cos B.
12.25=5.76+27.04-24.96 cos B, 12.25-5.76-27.04=-24.96 cos B. 20.54/24.96=cos B. 0.823=cos B.
B=34.61.
C=180-34.61-67.07=78.32.
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(D) Cosine Law - Examples
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(D) Cosine Law - Examples
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(D) Examples Cosine Law
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We can use these new trigonometric relationships in solving for
unknown sides and angles in acute triangles:
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ex 1. Find c in CDE if C = 56°, d = 4.7 and e = 8.5
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ex 2. Find G in GHJ if h = 5.9, g = 9.2 and j = 8.1
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ex 3. Solve CDE if D = 49°, e = 3.7 and c = 5.1
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