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Warm Up Find the area of each triangle with the given base and height.

Warm Up Find the area of each triangle with the given base and height. 1. b =10, h = 7 2. b = 8, h = 4.6. 35 units 2. 18.4 units 2. Solve each proportion. 3. 4. 10. 28.5. 5. In ∆ ABC , m  A = 122° and m  B =17°. What is the m  C ?. 41°. Objectives.

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Warm Up Find the area of each triangle with the given base and height.

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  1. Warm Up Find the area of each triangle with the given base and height. 1. b =10, h = 7 2.b = 8, h = 4.6 35 units2 18.4 units2 Solve each proportion. 3. 4. 10 28.5 5. In ∆ABC, mA = 122° and mB =17°. What is the mC ? 41°

  2. Objectives Determine the area of a triangle given side-angle-side information. Use the Law of Sines to find the side lengths and angle measures of a triangle.

  3. Area = ab sin C Example 1: Determining the Area of a Triangle Find the area of the triangle. Round to the nearest tenth. Write the area formula. Substitute 3 for a, 5 for b, and 40° for C. Use a calculator to evaluate the expression. ≈ 4.820907073 The area of the triangle is about 4.8 m2.

  4. Area = ac sin B Check It Out! Example 1 Find the area of the triangle. Round to the nearest tenth. Write the area formula. Substitute 8 for a, 12 for c, and 86° for B. Use a calculator to evaluate the expression. ≈ 47.88307441 The area of the triangle is about 47.9 m2.

  5. bc sin A = ac sin B = ab sin C bcsin A ac sin B ab sin C = = abc abc abc sin A = sin B = sin C b c a 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. Multiply each expression by 2. bc sin A = ac sin B = ab sin C Divide each expression by abc. Divide out common factors.

  6. The Law of Sines allows you to solve a triangle as long as you know either of the following: 1. Two angle measures and any side length–angle-angle-side (AAS) or angle-side-angle (ASA) information 2. Two side lengths and the measure of an angle that is not between them–side-side-angle (SSA) information

  7. 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. mD + mE + mF = 180° Triangle Sum Theorem. Substitute 33° for mD and 28° for mF. 33° + mE + 28° = 180° mE = 119° Solve for mE.

  8. sin F sin D sin F sin E = = d e f f sin 28° sin 28° sin 119° sin 33° = = e d 15 15 15 sin 33° 15 sin 119° d = e = sin 28° sin 28° d ≈ 17.4 e ≈ 27.9 Example 2A Continued Step 2 Find the unknown side lengths. Law of Sines. Substitute. Cross multiply. e sin 28° = 15 sin 119° d sin 28° = 15 sin 33° Solve for the unknown side.

  9. r Q 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. Triangle Sum Theorem mP = 180° – 36° – 39° = 105°

  10. 10 sin 36° 10 sin 39° q= r= ≈ 6.1 ≈ 6.5 sin 105° sin 105° r Q sin Q sin R sin P sin P = = p q p r sin 39° sin 36° sin 105° sin 105° = = r q 10 10 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. Law of Sines. Substitute.

  11. Check It Out! Example 2a Solve the triangle. Round to the nearest tenth. Step 1 Find the third angle measure. mH + mJ + mK = 180° Substitute 42° for mH and 107° for mJ. 42° + 107° + mK = 180° mK = 31° Solve for mK.

  12. sin J sin H sin H sin K = = h k h j sin 42° sin 107° sin 31° sin 42° = = k h 8.4 12 12 sin 42° 8.4 sin 31° h = k = sin 107° sin 42° h ≈ 8.4 k ≈ 6.5 Check It Out! Example 2a Continued Step 2 Find the unknown side lengths. Law of Sines. Substitute. Cross multiply. 8.4 sin 31° = k sin 42° h sin 107° = 12 sin 42° Solve for the unknown side.

  13. Check It Out! Example 2b Solve the triangle. Round to the nearest tenth. Step 1 Find the third angle measure. Triangle Sum Theorem mN = 180° – 56° – 106° = 18°

  14. sin M sin P sin N sin M = = sin 56° sin 106° n m m p = p 4.7 sin 106° sin 18° 1.5 sin 106° 4.7 sin 56° = m= p= ≈ 4.7 ≈ 4.0 m 1.5 sin 18° sin 106° Check It Out! Example 2b Solve the triangle. Round to the nearest tenth. Step 2 Find the unknown side lengths. Law of Sines. Substitute.

  15. When you use the Law of Sines to solve a triangle for which you know side-side-angle (SSA) information, zero, one, or two triangles may be possible. For this reason, SSA is called the ambiguous case.

  16. Remember! When one angle in a triangle is obtuse, the measures of the other two angles must be acute.

  17. C b a A B c Example 3: Art Application Determine the number of triangular banners that can be formed using the measurements a = 50, b = 20, and mA = 28°. Then solve the triangles. Round to the nearest tenth. Step 1 Determine the number of possible triangles. In this case, A is acute. Because b < a; only one triangle is possible.

  18. Example 3 Continued Step 2 Determine mB. Law of Sines Substitute. Solve for sin B.

  19. m B = Sin-1 Example 3 Continued Let B represent the acute angle with a sine of 0.188. Use the inverse sine function on your calculator to determine mB. Step 3 Find the other unknown measures of the triangle. Solve for mC. 28° + 10.8° + mC = 180° mC = 141.2°

  20. Example 3 Continued Solve for c. Law of Sines Substitute. Solve for c. c ≈ 66.8

  21. Check It Out! Example 3 Determine the number of triangles Maggie can form using the measurements a = 10 cm, b = 6 cm, and mA =105°. Then solve the triangles. Round to the nearest tenth. Step 1 Determine the number of possible triangles. In this case, A is obtuse. Because b < a; only one triangle is possible.

  22. Check It Out! Example 3 Continued Step 2 Determine mB. Law of Sines Substitute. Solve for sin B. sin B ≈ 0.58

  23. m B = Sin-1 Check It Out! Example 3 Continued Let B represent the acute angle with a sine of 0.58. Use the inverse sine function on your calculator to determine m B. Step 3 Find the other unknown measures of the triangle. Solve for mC. 105° + 35.4° + mC = 180° mC = 39.6°

  24. Check It Out! Example 3 Continued Solve for c. Law of Sines Substitute. Solve for c. c ≈ 6.6 cm

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