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 = 43Β°, a = 25, b = 28. Round side lengths to the nearest tenth and angle measures to the nearest degree.A is acute, and h = 28sin 43Β° 19.1Since a < b and a > h there are two different triangles.Solution 1: Find the acute B.
sinπ΅28
=π ππ 43 Β°
25
sinπ΅=28 π ππ 43 Β°
25
π΅=π ππβ1( 2 8 π ππ43 Β° 25 )
π΅β 50 Β°
sin 43 Β°25
βπ ππ87 Β°
π
π sin 43 Β° β 25 π ππ87 Β°
π β25 π ππ87 Β°
sin 43 Β°
π β 36.6
, C , and c 36.6
Example 4: Find two triangles for which A = 43Β°, 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 Β°π
βπ ππ 43 Β°
25
π β 4.5
π sin 43 Β° β 25 π ππ7 Β°
π β25 π ππ7 Β°sin 43 Β°
, C , and c 4.5
CB'B = B , so B' 180Β°β 50Β° 130Β°
C 180Β°β 43 Β°β 130Β° 7Β°
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 βABC. Round side lengths to the nearest tenth and angle measures to the nearest degree.Given a = 6, b =24, c = 20 find angle A.
π2=π2+π2 β2ππ cos π΄62=242 +202 β2 (24 ) (20 ) cos π΄
3 6=576+400 β960 cos π΄3 6=976 β960 cos π΄
β 940=β960 cos π΄940960
=cos π΄
πππ β 1 940960
=π΄
π΄β11.7 Β°
Example 6: Solve βABC. Round side lengths to the nearest tenth and angle measures to the nearest degree.
π2=π2 +π2 β2ππ cosπΆπ2=52+82 β2 (5 ) (8 ) cos 65 Β°
π2β 55.19π β7.4
sin π΄5
=π ππ65 Β°
7.4
sin π΄=5 π ππ65 Β°
7.4
π΄=π ππβ1( 5π ππ65 Β° 7.4 )
A β 38 Β°
, , 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
Heronβs Formula: SSS
1
2
Area s s a s b s c
s a b c
Example 7: Find the area of βXYZ.
π =12
(π₯+π¦+π§ )
π =67
π΄πππ=βπ (π βπ₯ ) (π β π¦ ) (π βπ§ )
π΄πππ=β683936π΄πππβ 827 ππ2
π =12
(45+51+38 )
π΄πππ=β67 (67 β 45 ) (67 β51 ) (67 β38 )
π₯=45 ππ .π¦=51 ππ .π§=38 ππ .
Example 8: Find the area of βGHJ to the nearest tenth.
π΄πππ=12πh sin π½
π=7 ππ h=10ππ π½=108 Β°
π΄πππ=12
(7 ) (10 ) sin 108 Β°
π΄πππβ 3 3.3ππ2
ββ1math!
4-7a Assignment: TX p298, 2-24 even
Unit 4 TestMonday 10/21
ββ1math!
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
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