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