math 160, precalculus - mwitt2010125/math30203/sinlaw.pdf · math 160, precalculus j. robert...

Law of SinesMATH 160, Precalculus

J. Robert Buchanan

Department of Mathematics

Fall 2011

J. Robert Buchanan Law of Sines

Objectives

In this lesson we will learn to:use the Law of Sines to solve oblique triangles (AAS, ASA,and SSA),find the areas of oblique triangles,use the the Law of Sines to model and solve real-worldproblems.

Oblique Triangles

Oblique triangles are triangles having no right angles.

Solving an Oblique Triangle

Solving a triangle involves determining to the lengths of itsthree sides and the measures of its three angles.

To solve a triangle we must know the length of at least one sideand any two other measures of the triangle:

1 Any side and two angles (AAS or ASA).2 Two sides and the angle opposite one of them (SSA).3 Three sides (SSS).4 Two sides and their included angle.

Today we will explore the first two cases.

Law of Sines

Law of SinesIf ABC is a triangle with sides a, b, and c, then

asin A

sin B=

csin C

orsin A

sin Bb

=sin C

Acute triangle Obtuse triangle

Example

Solve the following triangle (not drawn to scale).

35° 40°

This is the ASA case.

C = 180◦ − 35◦ − 40◦ = 105◦

sin Csin A =

10 sin 35◦

sin 105◦ ≈ 5.94

sin Csin B =

10 sin 40◦

sin 105◦ ≈ 6.65

Example

Solve the following triangle (not drawn to scale).

35° 40°

This is the ASA case.

C = 180◦ − 35◦ − 40◦ = 105◦

sin Csin A =

10 sin 35◦

sin 105◦ ≈ 5.94

sin Csin B =

10 sin 40◦

sin 105◦ ≈ 6.65

Example

A flagpole at a right angle to the horizontal is located on a slopethat makes an angle of 12◦ with the horizontal. The flagpole’sshadow is 16 meters long and points directly up the slope. Theangle of elevation from the tip of the shadow to the sun is 20◦.Find the height of the flagpole.

This is the ASA situation. Let A = 78◦, B = 32◦, and c = 16meters.

C = 180◦ − 78◦ − 32◦ = 70circ

sin Csin B =

16 sin 70◦

sin 32◦ ≈ 9.02 meters

Example

A flagpole at a right angle to the horizontal is located on a slopethat makes an angle of 12◦ with the horizontal. The flagpole’sshadow is 16 meters long and points directly up the slope. Theangle of elevation from the tip of the shadow to the sun is 20◦.Find the height of the flagpole.This is the ASA situation. Let A = 78◦, B = 32◦, and c = 16meters.

C = 180◦ − 78◦ − 32◦ = 70circ

sin Csin B =

16 sin 70◦

Example

A flagpole at a right angle to the horizontal is located on a slopethat makes an angle of 12◦ with the horizontal. The flagpole’sshadow is 16 meters long and points directly up the slope. Theangle of elevation from the tip of the shadow to the sun is 20◦.Find the height of the flagpole.This is the ASA situation. Let A = 78◦, B = 32◦, and c = 16meters.

C = 180◦ − 78◦ − 32◦ = 70circ

sin Csin B =

16 sin 70◦

SSA: Ambiguous Case

If we know two side lengths and the measure of the angleopposite one of the sides, three possibles outcomes maypresent themselves when solving the oblique triangle.

No such triangle exists.One such triangle exists.Two distinct triangles exist.

Example

Use the Law of Sines to solve the triangle, if there is a solution.If two solutions exist, find them both.

A = 110◦, a = 125, b = 200.

sin Aa

=sin B

sin B =b sin A

=200 sin 110◦

125= 1.50351 > 1

No solution.

Example

A = 110◦, a = 125, b = 200.

sin Aa

=sin B

sin B =b sin A

=200 sin 110◦

125= 1.50351 > 1

No solution.

Example

A = 76◦, a = 34, b = 21.

sin Aa

=sin B

sin B =b sin A

=21 sin 76◦

34≈ 0.5993

B ≈ 36.82◦

C ≈ 67.18◦

sin Asin C ≈ 34 sin 67.18◦

sin 76◦ ≈ 32.30

Example

A = 76◦, a = 34, b = 21.

sin Aa

=sin B

sin B =b sin A

=21 sin 76◦

34≈ 0.5993

B ≈ 36.82◦

C ≈ 67.18◦

sin Asin C ≈ 34 sin 67.18◦

sin 76◦ ≈ 32.30

Example

A = 25◦, a = 9.5, b = 22.

sin Aa

=sin B

sin B =b sin A

=22 sin 25◦

9.5≈ 0.978695

B ≈ 78.15◦ or B ≈ 101.85◦

C ≈ 76.85◦ or C ≈ 53.15◦

c ≈ 9.5 sin 76.85◦

sin 25◦ ≈ 21.89 or c ≈ 9.5 sin 53.15◦

sin 25◦ ≈ 17.99

Example

A = 25◦, a = 9.5, b = 22.

sin Aa

=sin B

sin B =b sin A

=22 sin 25◦

9.5≈ 0.978695

B ≈ 78.15◦ or B ≈ 101.85◦

C ≈ 76.85◦ or C ≈ 53.15◦

c ≈ 9.5 sin 76.85◦

sin 25◦ ≈ 21.89 or c ≈ 9.5 sin 53.15◦

sin 25◦ ≈ 17.99

Area of an Oblique Triangle

AreaThe area of any triangle is one-half the product of the lengths oftwo sides times the sine of their included angle.

Area =12

bc sin A =12

ac sin B =12

ab sin C

Acute triangle Obtuse triangle

Example

Find the area of the triangle having the following angle andsides.

B = 130◦, a = 62, c = 20.

Area =12

ac sin B =12(62)(20) sin 130◦ = 474.948

Example

Find the area of the triangle having the following angle andsides.

B = 130◦, a = 62, c = 20.

Area =12

ac sin B =12(62)(20) sin 130◦ = 474.948

Application

The circular arc of a railroad curve has a chord of length 3000feet corresponding to a central angle of 40◦.

Find the radius r of the circular arc.

rsin 70◦ =

3000sin 40◦

r =3000 sin 70◦

sin 40◦ ≈ 4385.71 feet

Find the length s of the circular arc.

The central angle of 40◦ is equivalent to 2π/9 radians

s = rθ = 4385.71(

)= 3061.8 feet

Application

rsin 70◦ =

3000sin 40◦

r =3000 sin 70◦

sin 40◦ ≈ 4385.71 feet

Find the length s of the circular arc.

The central angle of 40◦ is equivalent to 2π/9 radians

s = rθ = 4385.71(

)= 3061.8 feet

Application

rsin 70◦ =

3000sin 40◦

r =3000 sin 70◦

sin 40◦ ≈ 4385.71 feet

Find the length s of the circular arc.The central angle of 40◦ is equivalent to 2π/9 radians

s = rθ = 4385.71(

)= 3061.8 feet

Homework

Read Section 5.6.Exercises: 1, 5, 9, 13, . . . , 49, 53

math 160, precalculus - mwitt2010125/math30203/sinlaw.pdf · math 160, precalculus j. robert...

Documents