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Objectives. Use trig. to find the area of triangles. Use the Law of Sines to find the side lengths and angle measures of a triangle. Notes #1-3. Find the area of the triangle. Round to the nearest tenth. 2. Solve the triangle. - PowerPoint PPT Presentation
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Holt Algebra 2
13-5 The Law of Sines
Use trig. to find the area of triangles.
Use the Law of Sines to find the side lengths and angle measures of a triangle.
Objectives
Holt Algebra 2
13-5 The Law of Sines
Notes #1-3
1. Find the area of the triangle. Round to the nearest tenth.
2. Solve the triangle.
3. Triangular banners can be formed using the measurements a = 48, b = 28, and mA = 35°. Solve the triangle (nearest tenth).
Holt Algebra 2
13-5 The Law of Sines
Example 1: Determining the Area of a Triangle
Find the area of the triangle. Round to the nearest tenth.
Area = ab sin C
≈ 4.82
Write the area formula.
Substitute 3 for a, 5 for b, and 40° for C.
Use a calculator to evaluate the expression (round).
Holt Algebra 2
13-5 The Law of Sines
The area of ∆ABC is equal to bc sin A or ac sin B or ab sin C. By setting these expressions equal to each other, you can derive the Law of Sines.
bc sin A = ac sin B = ab sin C
bc sin A = ac sin B = ab sin C
bc sin A ac sin B ab sin C abc abc abc
= =
sin A = sin B = sin C a b c
Multiply each expression by 2.
Divide each expression by abc.
Divide out common factors.
Holt Algebra 2
13-5 The Law of Sines
Holt Algebra 2
13-5 The Law of Sines
Example 2A: Using the Law of Sines for AAS and ASA
Solve the triangle. Round to the nearest tenth.
Step 1. Find the third angle measure.
33° + mE + 28° = 180°
mE = 119°
Substitute 33° for mD and 28° for mF.
Solve for mE.
Holt Algebra 2
13-5 The Law of Sines
Example 2A Continued
Step 2 Find the unknown side lengths.
sin D sin Fd f
=sin E sin F
e f=
sin 33° sin 28°d 15=
sin 119° sin 28°e 15=
d sin 28° = 15 sin 33° e sin 28° = 15 sin 119°
d = 15 sin 33°sin 28°
d ≈ 17.4
e = 15 sin 119°sin 28°
e ≈ 27.9Solve for the
unknown side.
Law of Sines.
Substitute.
Crossmultiply.
Holt Algebra 2
13-5 The Law of Sines
Example 2B: Using the Law of Sines for AAS and ASA
Solve the triangle. Round to the nearest tenth.
Step 1 Find the third angle measure.
mP = 180° – 36° – 39° = 105° Triangle Sum Theorem
Q
r
Holt Algebra 2
13-5 The Law of Sines
Example 2B: Using the Law of Sines for AAS and ASA
Solve the triangle. Round to the nearest tenth.
Step 2 Find the unknown side lengths.
sin P sin Qp q= sin P sin R
p r=Law of Sines.
sin 105° sin 36°10 q= sin 105° sin 39°
10 r=Substitute.
q = 10 sin 36°sin 105°
≈ 6.1 r = 10 sin 39°sin 105°
≈ 6.5
Q
r
Holt Algebra 2
13-5 The Law of Sines
Example 3: Art Application
Triangular banners can be formed using the measurements a = 50, b = 20, and mA = 28°. Solve the triangle (nearest tenth).
Step 1 Determine mB.
m B = Sin-1
Holt Algebra 2
13-5 The Law of Sines
Example 3 Continued
Solve for c.
c ≈ 66.8
Solve for c.
Step 3 Find the other unknowns in the triangle.
28° + 10.8° + mC = 180°
mC = 141.2°
Holt Algebra 2
13-5 The Law of Sines
Notes #1-3
1. Find the area of the triangle. Round to the nearest tenth.
17.8 ft2
2. Solve the triangle. Round to the nearest tenth.
a 32.2; b 22.0; mC = 133.8°
Holt Algebra 2
13-5 The Law of Sines
Example 3: Art Application
Triangular banners can be formed using the measurements a = 48, b = 28, and mA = 35°. Solve the triangle (nearest tenth).
Holt Algebra 2
13-5 The Law of Sines
Notes #3
3. Determine the number of triangular quilt pieces that can be formed by using the measurements a = 14 cm, b = 20 cm, and mA = 39°. Solve each triangle. Round to the nearest tenth. 2;
c1 21.7 cm;mB1 ≈ 64.0°;mC1 ≈ 77.0°;
c2 ≈ 9.4 cm;mB2 ≈ 116.0°;mC2 ≈ 25.0°
Holt Algebra 2
13-5 The Law of Sines
Holt Algebra 2
13-5 The Law of Sines
Solving a Triangle Given a, b, and mA