8-6 The Law of Sines and Law of Cosines

You used trigonometric ratios to solve right triangles.

• Use the Law of Sines to solve triangles.

• Use the Law of Cosines to solve triangles.

Law of Sines• The Law of Sines can be used to find side lengths

and angle measures for any triangle (nonright triangles).

• You can use the Law of Sines to solve a triangle if you know the measures of two angles and any side (AAS or ASA)

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Find p. Round to the nearest tenth.

We are given measures of two angles and a nonincluded side, so use the Law of Sines to write a proportion (AAS).

Law of Sines

Use a calculator.

Divide each side by sin

Cross Products Property

Answer: p ≈ 4.8

Law of Sines (AAS or ASA)

A. 4.6

B. 29.9

C. 7.8

D. 8.5

Find c to the nearest tenth.

Find x. Round to the nearest tenth.6

x 57°

Law of Sines (ASA)

Law of Sines

mB = 50, mC = 73, c = 6

6 sin 50 = x sin 73 Cross Products Property

Divide each side by sin 73.

Answer: x ≈ 4.8

4.8 = x Use a calculator.

A. 8

B. 10

C. 12

D. 14

Find x. Round to the nearest degree.

43°

x

Law of Cosines

You can use the Law of Cosines to solve a triangle if you know the measures of two sides and the included angle (SAS)

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Law of Cosines (SAS)Find x. Round to the nearest tenth.Use the Law of Cosines since the measures of two sides and the included angle are known.

Answer: x ≈ 18.9

Simplify.

Take the square root of each side.

Law of Cosines

Use a calculator.

A. 25.1

B. 44.5

C. 22.7

D. 21.1

Find r if s = 15, t = 32, and mR = 40. Round to the nearest tenth.

Law of Cosines (SSS)

Find mL. Round to the nearest degree.

Law of Cosines

Simplify.

Answer: mL ≈ 49

Solve for L.

Use a calculator.

Subtract 754 from each side.Divide each side by –270.

A. 44°

B. 51°

C. 56°

D. 69°

Find mP. Round to the nearest degree.

8-6 Assignment, day 1

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8-6 The Law of Sines and Law of Cosines, day 2

You used trigonometric ratios to solve right triangles.

• Use the Law of Sines to solve triangles.

• Use the Law of Cosines to solve triangles.

Solve a Triangle

When solving right triangles, you can use sine, cosine, or tangent.

When solving other triangles, you can use the Law of Sines or the Law of Cosines, depending on what information is given AAS, ASA for sinesSAS, SSS for cosines)

AIRCRAFT From the diagram of the plane shown, determine the approximate width of each wing. Round to the nearest tenth meter.

Cross products

Law of Sines

Use the Law of Sines to find KJ.

Answer: The width of each wing is about 16.9 meters.

Simplify.

Divide each side by sin .

Cross products

A. 93.5 in.

B. 103.5 in.

C. 96.7 in.

D. 88.8 in.

The rear side window of a station wagon has the shape shown in the figure. Find the perimeter of the window if the length of DB is 31 inches. Round to the nearest tenth.

Solve a TriangleSolve triangle PQR. Round to the nearest degree.Since the measures of three sides are given (SSS), use the Law of Cosines to find mP.

p2 = r2 + q2 – 2pq cos P Law of Cosines

82 = 92 + 72 – 2(9)(7) cos P p = 8, r = 9, and q = 764 = 130 – 126 cos P Simplify.

–66 = –126 cos P Subtract 130 from each side.

Divide each side by –126.

Use the inverse cosine ratio.

Use a calculator.

Use the Law of Sines to find mQ.

Law of Sines

Multiply each side by 7.Use the inversesine ratio.Use a calculator.

mP ≈ 58, p = 8,q = 7

Answer: Therefore, mP ≈ 58; mQ ≈ 48 andmR ≈ 74.

By the Triangle Angle Sum Theorem, mR ≈ 180 – (58 + 48) or 74.

A. mR = 82, mS = 58, mT = 40

B. mR = 58, mS = 82, mT = 40

C. mR = 82, mS = 40, mT = 58

D. mR = 40, mS = 58, mT = 82

Solve ΔRST. Round to the nearest degree.

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SAS

AAS

ASA

8-6 Assignment day 2

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