Holt Geometry Law of Cosines 10.5. Holt Geometry The Law of Sines cannot be used to solve every...
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Transcript of Holt Geometry Law of Cosines 10.5. Holt Geometry The Law of Sines cannot be used to solve every...
Holt Geometry
Law of Cosines 10.5
Holt Geometry
The Law of Sines cannot be used to solve every triangle. Instead, you can apply the Law of Cosines.You can use the Law of Cosines to solve a triangle if you are given• two side lengths and the included angle measure
(SAS) or• three side lengths (SSS).
Holt Geometry
Law of Cosines
Standard Form
2 2 2 2 cosa b c bc A
Alternative Form2 2 2
cos2
b c aAbc
2 2 2 2 cosb a c ac B 2 2 2
cos2
a c bBac
2 2 2 2 cosc a b ab C 2 2 2
cos2
a b cCab
Holt Geometry
The angle referenced in the Law of Cosines is across the equal sign from its corresponding side.
Helpful Hint
Do not round your answer until the final step of the computation. If a problem has multiple steps, store the calculated answers to each part in your calculator.
Helpful Hint
Holt Geometry
Find the measure of XZ.
XZ2 = XY2 + YZ2 – 2(XY)(YZ)cos Y
XZ2 = 352 + 302 – 2(35)(30)cos 110°
XZ2 2843.2423
XZ 53.3
Law of Cosines
Substitution.
Simplify.
Take the square root of both sides.
Law of Cosines – SAS
Holt Geometry
Find mT.
RS2 = RT2 + ST2 – 2(RT)(ST)cos T
72 = 132 + 112 – 2(13)(11)cos T
49 = 290 – 286 cosT
–241 = –286 cosT
Law of Cosines
Substitution.
Simplify.
Subtract 290 from both sides.
Law of Cosines – SSS
Division.
Use inverse cosine to find mT.
Holt Geometry
Find the measure of DE.
DE2 = EF2 + DF2 – 2(EF)(DF)cos F
DE2 = 182 + 162 – 2(18)(16)cos 21°
DE2 42.2577
DE 6.5
Law of Cosines
Substitution.
Simplify.
Take the square root of both sides.
Law of Cosines – SAS
Holt Geometry
Find mK.
JL2 = LK2 + KJ2 – 2(LK)(KJ)cos K
82 = 152 + 102 – 2(15)(10)cos K
64 = 325 – 300 cosK
–261 = –300 cosK
Law of Cosines
Substitution.
Simplify.
Subtract 325 both sides.
Law of Cosines – SSS
Division.
Use inverse cosine to find mK.
Holt Geometry
Find the measure of YZ.
YZ2 = XY2 + XZ2 – 2(XY)(XZ)cos X
YZ2 = 102 + 42 – 2(10)(4)cos 34°
YZ2 49.6770
YZ 7.0
Law of Cosines
Substitution.
Simplify.
Take the square root of both sides.
Law of Cosines – SAS
Holt Geometry
Find mR.
PQ2 = PR2 + RQ2 – 2(PR)(RQ)cos R
9.62 = 5.92 + 10.52 – 2(5.9)(10.5)cos R
92.16 = 145.06 – 123.9cosR
–52.9 = –123.9 cosR
Law of Cosines
Substitution.
Simplify.
Subtract 145.06 both sides.
Law of Cosines – SSS
Solve for cosR.
Use inverse cosine to find mR.
Holt Geometry
Solve the following triangle.
Law of Cosines – SAS
BC2 = AB2 + AC2 – 2(AB)(AC)cos A BC2 = 3.92 + 3.12 – 2(3.9)(3.1)cos 45° BC2 7.7222
BC 2.8 mi
Law of CosinesSubstitution.Simplify.Take the square root of both sides.
Law of Sines
Substitution.
Multiply both sides by 3.9.
Use inverse sine to find mC.
Holt Geometry
An engineer suggested using a cable attached from the top of the tower to a point 31 m from the base. How long would this cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree.
31 m
Law of Cosines – SAS
AC2 = AB2 + BC2 – 2(AB)(BC)cos B
AC2 = 312 + 562 – 2(31)(56)cos 100°
AC2 4699.9065
AC 68.6 m
Law of Cosines
Substitution.
Simplify.
Take the square root of both sides.
Holt Geometry
Find the measure of the angle the cable would make with the ground.
Law of Sines
Substitution.
Multiply both sides by 56.
Use inverse sine to find mA.
31 m
Holt Geometry
Use ΔABC for a – c. Round lengths to the nearest tenth and angle measures to the nearest degree.
a. mB = 20°, mC = 31° and b = 210. Find a.
b. a = 16, b = 10, and mC = 110°. Find c.
c. a = 20, b = 15, and c = 8.3. Find mA.
477.2
21.6
115°
Holt Geometry
An observer in tower A sees a fire 1554 ft away at an angle of depression of 28°. To the nearest foot, how far is the fire from an observer in tower B? To the nearest degree, what is the angle of depression to the fire from tower B?
1212 ft; 37°
Holt Geometry
AssignmentGeometry:
10.1
Law of Cosines
Holt Geometry
Heron’s Area Formula
Given any triangle with sides of lengths a, b, and c, the area of the triangle is given by
Area ( )( )( )s s a s b s c
Find the area of the triangle.
11.5 82
52
10a b cs
Area ( 5)(11.5 11.5 11.5 11.8)( 10)5 5
108
where .2
a b cs
19.8 square units
Holt Geometry
EXAMPLE 4
The intersection of three streets forms a piece of land called a traffic triangle. Find the area of the traffic triangle shown.
Find the semiperimeter s.
s = (a + b + c )12
12= (170 + 240 + 350) = 380
Area = s (s – a) (s – b) (s – c)
18,300
The area of the traffic triangle is about 18,300 square yds.
380 (380 – 170) (380 – 240) (380 – 350)=Area
Holt Geometry
GUIDED PRACTICE
Find the area of ABC.
Find the semiperimeter s.
s = (a + b + c )12
12= (5 + 8 + 11) = 12
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.
18.312 (12 – 8) (12 – 11) (12 – 5)=Area
Holt Geometry
GUIDED PRACTICE
Find the area of ABC.
Find the semiperimeter s.
s = (a + b + c )12
12= (4 + 9+ 7) = 10
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.
13.410 (10– 4) (10 – 9) (10 – 7)=Area
Holt Geometry
Find the area of ABC.
Find the semiperimeter s.
s = (a + b + c )12
12= (15 + 23+ 127) = 25
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.
80.625 (25– 15) (25 – 23) (25 – 12)=Area
Holt Geometry
AssignmentGeometry:
10.5
Heron’s Law
