use trig. to find the area of triangles
DESCRIPTION
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 PresentationTRANSCRIPT
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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
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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).
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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).
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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.
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Holt Algebra 2
13-5 The Law of Sines
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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.
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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.
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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
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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
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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
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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°
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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°
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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).
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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°
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Holt Algebra 2
13-5 The Law of Sines
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Holt Algebra 2
13-5 The Law of Sines
Solving a Triangle Given a, b, and mA