Holt Geometry 8-5 Law of Sines and Law of Cosines 8-5 Law of Sines and Law of Cosines Holt Geometry...

Holt Geometry

8-5 Law of Sines and Law of Cosines8-5 Law of Sines and Law of Cosines

Holt Geometry

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Geometry

8-5 Law of Sines and Law of Cosines

Warm Up1. What is the third angle measure in a triangle with

angles measuring 65° and 43°?

Find each value. Round trigonometric

ratios to the nearest hundredth and angle

measures to the nearest degree.

2. sin 73° 3. cos 18° 4. tan 82°

5. sin-1 (0.34) 6. cos-1 (0.63) 7. tan-1 (2.75)

72°

0.96 0.95 7.12

20° 51° 70°

Holt Geometry


Use the Law of Sines and the Law of Cosines to solve triangles.

Objective

Holt Geometry


In this lesson, you will learn to solve any triangle.To do so, you will need to calculate trigonometricratios for angle measures up to 180°. You can use a calculator to find these values.

Holt Geometry


Example 1: Finding Trigonometric Ratios for Obtuse Angles

Use your calculator to find each trigonometric ratio. Round to the nearest hundredth.

A. tan 103° B. cos 165° C. sin 93°

tan 103° –4.33 cos 165° –0.97 sin 93° 1.00

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Check It Out! Example 1

Use a calculator to find each trigonometric ratio. Round to the nearest hundredth.

a. tan 175°

tan 175° –0.09

b. cos 92° c. sin 160°

cos 92° –0.03 sin 160° 0.34

Holt Geometry


You can use the altitude of a triangle to find a relationship between the triangle’s side lengths.

In ∆ABC, let h represent the length of the altitude from C to

From the diagram, ,

and

By solving for h, you find that h = b sin A and h = a

sin B. So b sin A = a sin B, and .

You can use another altitude to show that these

ratios equal

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You can use the Law of Sines to solve a triangle if you are given• two angle measures and any side length

(ASA or AAS) or• two side lengths and a non-included angle measure

(SSA).

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Example 2A: Using the Law of Sines

Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree.

FG

Law of Sines

Substitute the given values.

Cross Products Property

Divide both sides by sin 39.

FG sin 39° = 40 sin 32°

Holt Geometry


Example 2B: Using the Law of Sines

Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree.mQ

Law of Sines


Multiply both sides by 6.

Use the inverse sine function to find mQ.

Holt Geometry


Check It Out! Example 2a


NP

Law of Sines



Divide both sides by sin 39°.

NP sin 39° = 22 sin 88°

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Check It Out! Example 2b


mL

Law of Sines



Use the inverse sine function to find mL.

10 sin L = 6 sin 125°

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Check It Out! Example 2c


mX

Law of Sines



Use the inverse sine function to find mX.

7.6 sin X = 4.3 sin 50°

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Check It Out! Example 2d


AC

mA + mB + mC = 180°

mA + 67° + 44° = 180°

mA = 69°

Prop of ∆.


Simplify.

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Check It Out! Example 2D Continued

Law of Sines



Divide both sides by sin 69°.


AC sin 69° = 18 sin 67°

Holt Geometry


The Law of Sines cannot be used to solve every triangle. If you know two side lengths and the included angle measure or if you know all three side lengths, you cannot use the Law of Sines. Instead, you can apply the Law of Cosines.

Holt Geometry


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).

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The angle referenced in the Law of Cosines is across the equal sign from its corresponding side.

Helpful Hint

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Example 3A: Using the Law of Cosines


XZ

XZ2 = XY2 + YZ2 – 2(XY)(YZ)cos Y

= 352 + 302 – 2(35)(30)cos 110°

XZ2 2843.2423

XZ 53.3

Law of CosinesSubstitute the

given values.

Simplify.

Find the square root of both sides.

Holt Geometry


Example 3B: Using the Law of Cosines


mT

RS2 = RT2 + ST2 – 2(RT)(ST)cos T

72 = 132 + 112 – 2(13)(11)cos T

49 = 290 – 286 cosT

–241 = –286 cosT


given values.

Simplify.

Subtract 290 both sides.

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Example 3B Continued


mT

–241 = –286 cosT

Solve for cosT.

Use the inverse cosine function to find mT.

Holt Geometry


Check It Out! Example 3a


DE

DE2 = EF2 + DF2 – 2(EF)(DF)cos F

= 182 + 162 – 2(18)(16)cos 21°

DE2 42.2577

DE 6.5


given values.

Simplify.


Holt Geometry


Check It Out! Example 3b


mK

JL2 = LK2 + KJ2 – 2(LK)(KJ)cos K

82 = 152 + 102 – 2(15)(10)cos K

64 = 325 – 300 cosK

–261 = –300 cosK


given values.

Simplify.

Subtract 325 both sides.

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Check It Out! Example 3b Continued


mK

–261 = –300 cosK

Solve for cosK.

Use the inverse cosine function to find mK.

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Check It Out! Example 3c


YZ

YZ2 = XY2 + XZ2 – 2(XY)(XZ)cos X

= 102 + 42 – 2(10)(4)cos 34°

YZ2 49.6770

YZ 7.0


given values.

Simplify.


Holt Geometry


Check It Out! Example 3d


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


given values.

Simplify.

Subtract 145.06 both sides.

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Check It Out! Example 3d Continued


mR

–52.9 = –123.9 cosR

Solve for cosR.

Use the inverse cosine function to find mR.

Holt Geometry


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


Example 4: Sailing Application

A sailing club has planned a triangular racecourse, as shown in the diagram. How long is the leg of the race along BC? How many degrees must competitors turn at point C? Round the length to the nearest tenth and the angle measure to the nearest degree.

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Example 4 Continued

Step 1 Find BC.

BC2 = AB2 + AC2 – 2(AB)(AC)cos A

= 3.92 + 3.12 – 2(3.9)(3.1)cos 45°

BC2 7.7222

BC 2.8 mi

Law of Cosines


Simplify.


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Step 2 Find the measure of the angle through which competitors must turn. This is mC.

Example 4 Continued

Law of Sines


Multiply both sides by 3.9.

Use the inverse sine function to find mC.

Holt Geometry


Check It Out! Example 4

What if…? Another 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

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Check It Out! Example 4 Continued

Step 1 Find the length of the cable.

AC2 = AB2 + BC2 – 2(AB)(BC)cos B

= 312 + 562 – 2(31)(56)cos 100°

AC2 4699.9065

AC 68.6 m

Law of Cosines


Simplify.


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Step 2 Find the measure of the angle the cable would make with the ground.

Check It Out! Example 4 Continued

Law of Sines


Multiply both sides by 56.

Use the inverse sine function to

find mA.

Holt Geometry


Lesson Quiz: Part I

Use a calculator to find each trigonometric ratio. Round to the nearest hundredth.

1. tan 154°

2. cos 124°

3. sin 162°

–0.49

–0.56

0.31

Holt Geometry


Use ΔABC for Items 4–6. Round lengths to the nearest tenth and angle measures to the nearest degree.

4. mB = 20°, mC = 31° and b = 210. Find a.

5. a = 16, b = 10, and mC = 110°. Find c.

6. a = 20, b = 15, and c = 8.3. Find mA.

Lesson Quiz: Part II

477.2

21.6

115°

Holt Geometry


Lesson Quiz: Part III

7. 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 8-5 Law of Sines and Law of Cosines 8-5 Law of Sines and Law of Cosines Holt Geometry...

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