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Warm-Up 10/151.
H
PSAT Tomorrow 10/16, you will need to bring your own calculator.
Rigor:You will learn how to use the Law of
Sines and Cosines to solve right triangle problems.
Relevance:You will be able to use the Law of Sines
and Cosines to solve real world problems.
4-7b Law of Sines and the Law of Cosines
Cases:
1 ASA or SAA2 SSA3 SAS4 SSS
LAW OF SINES
LAW OF COSINES
Example 4: Find two triangles for which A = 43°, a = 25, b = 28. Round side lengths to the nearest tenth and angle measures to the nearest degree.A is acute, and h = 28sin 43° 19.1Since a < b and a > h there are two different triangles.Solution 1: Find the acute B.
sin𝐵28
=𝑠𝑖𝑛 43 °
25
sin𝐵=28 𝑠𝑖𝑛 43 °
25
𝐵=𝑠𝑖𝑛−1( 2 8 𝑠𝑖𝑛43 ° 25 )
𝐵≈ 50 °
sin 43 °25
≈𝑠𝑖𝑛87 °
𝑐
𝑐 sin 43 ° ≈ 25 𝑠𝑖𝑛87 °
𝑐 ≈25 𝑠𝑖𝑛87 °
sin 43 °
𝑐 ≈ 36.6
, C , and c 36.6
Example 4: Find two triangles for which A = 43°, a = 25, b = 28. Round side lengths to the nearest tenth and angle measures to the nearest degree.Solution 2: Find the obtuse B'.
sin 7 °𝑐
≈𝑠𝑖𝑛 43 °
25
𝑐 ≈ 4.5
𝑐 sin 43 ° ≈ 25 𝑠𝑖𝑛7 °
𝑐 ≈25 𝑠𝑖𝑛7 °sin 43 °
, C , and c 4.5
CB'B = B , so B' 180°– 50° 130°
C 180°– 43 °– 130° 7°
Law of Cosines:2 2 2
2 2 2
2 2 2
2 cos
2 cos
2 cos
a b c bc A
b a c ac B
c b a ab C
Example 5: Solve ∆ABC. Round side lengths to the nearest tenth and angle measures to the nearest degree.Given a = 6, b =24, c = 20 find angle A.
𝑎2=𝑏2+𝑐2 −2𝑏𝑐 cos 𝐴62=242 +202 −2 (24 ) (20 ) cos 𝐴
3 6=576+400 −960 cos 𝐴3 6=976 −960 cos 𝐴
− 940=−960 cos 𝐴940960
=cos 𝐴
𝑐𝑜𝑠− 1 940960
=𝐴
𝐴≈11.7 °
Example 6: Solve ∆ABC. Round side lengths to the nearest tenth and angle measures to the nearest degree.
𝑐2=𝑎2 +𝑏2 −2𝑎𝑏 cos𝐶𝑐2=52+82 −2 (5 ) (8 ) cos 65 °
𝑐2≈ 55.19𝑐 ≈7.4
sin 𝐴5
=𝑠𝑖𝑛65 °
7.4
sin 𝐴=5 𝑠𝑖𝑛65 °
7.4
𝐴=𝑠𝑖𝑛−1( 5𝑠𝑖𝑛65 ° 7.4 )
A ≈ 38 °
, , and c
Law of Cosines:
2 2 2
2 2 2
2 2 2
cos2
cos2
cos2
b c aA
bc
a c bB
ac
a b cC
ab
Heron’s Formula: SSS
1
2
Area s s a s b s c
s a b c
Example 7: Find the area of ∆XYZ.
𝑠=12
(𝑥+𝑦+𝑧 )
𝑠=67
𝐴𝑟𝑒𝑎=√𝑠 (𝑠−𝑥 ) (𝑠− 𝑦 ) (𝑠−𝑧 )
𝐴𝑟𝑒𝑎=√683936𝐴𝑟𝑒𝑎≈ 827 𝑖𝑛2
𝑠=12
(45+51+38 )
𝐴𝑟𝑒𝑎=√67 (67 − 45 ) (67 −51 ) (67 −38 )
𝑥=45 𝑖𝑛 .𝑦=51 𝑖𝑛 .𝑧=38 𝑖𝑛 .
Example 8: Find the area of ∆GHJ to the nearest tenth.
𝐴𝑟𝑒𝑎=12𝑔h sin 𝐽
𝑔=7 𝑐𝑚 h=10𝑐𝑚 𝐽=108 °
𝐴𝑟𝑒𝑎=12
(7 ) (10 ) sin 108 °
𝐴𝑟𝑒𝑎≈ 3 3.3𝑐𝑚2
√−1math!
4-7a Assignment: TX p298, 2-24 even
Unit 4 TestMonday 10/21
√−1math!
4-7a Assignment: TX p298, 2-24 even
Assignment for Wednesday:4-7b Assignment: TX p298-299, 26-50 even
Unit 4 TestMonday 10/21