Five-Minute Check (over Lesson 8–5) Mathematical Practices Then/Now New Vocabulary Key Concept: Area of a Triangle Example 1: Find the Area of a Triangle Key Concept: Law of Sines Example 2: Solve a Triangle Given Two Angles and a Side Key Concept: Possible Triangles in SSA Case Example 3: Solve a Triangle Given Two Sides and an Angle Example 4: Real-World Example: Use the Law of Sines to Solve a Problem Lesson Menu

Find the exact values of sin , cos , and tan  if the terminal side of  in standard position contains the point at (3, –4). A. B. C. D. 5-Minute Check 1

Find the exact value of cos 150°. A. ans B. ans C. ans D. ans 5-Minute Check 2

Find the exact value of csc . B. 1 C. –1 D. –2 5-Minute Check 3

The tension in a pendulum’s string is modeled by T = 20 cos  The tension in a pendulum’s string is modeled by T = 20 cos . Determine the degree measure of the angle for T = 0, 10, and 20. A. 0°, 45°, 90° B. 90°, 45°, 0° C. 90°, 60°, 45° D. 90°, 60°, 0° 5-Minute Check 4

Which of the following trigonometric functions are positive in quadrant III? A. sine B. cosine C. tangent D. secant 5-Minute Check 5

Mathematical Practices 1 Make sense of problems and persevere in solving them. 3 Construct viable arguments and critique the reasoning of others. MP

You found side lengths and angle measures of right triangles. Find the area of a triangle using two sides and an included angle. Use the Law of Sines to solve triangles. Then/Now

Law of Sines solving a triangle ambiguous case Vocabulary

Concept

In this triangle, b = 6, c = 3, and A = 25°. Use the formula Find the Area of a Triangle In this triangle, b = 6, c = 3, and A = 25°. Use the formula Area Area formula Replace b with 6, c with 3, and A with 25°. Use a calculator. Answer: The area is about 3.8 square centimeters. Example 1

Find the area of ΔABC to the nearest tenth. A. 2 cm2 B. 5 cm2 C. 7 cm2 D. 14 cm2 Example 1

Concept

nearest tenth if necessary. Solve a Triangle Given Two Angles and a Side Round to the nearest tenth if necessary. You are given the measures of two angles and a side. First, find the measure of the third angle. 100° + 53° + m A =180° The sum of the angle measures of a triangle is 180°. m A = 27° 180 – (100 + 53) = 27 Example 2

Use the Law of Sines to find a and c. Solve a Triangle Given Two Angles and a Side Use the Law of Sines to find a and c. Law of Sines Replace A with 27°, B with 53°, and b with 9. Solve for a. Use a calculator. Example 2

Replace B with 53°, C with 100°, and b with 9. Solve a Triangle Given Two Angles and a Side Law of Sines Replace B with 53°, C with 100°, and b with 9. Solve for c. Use a calculator. Example 2

Solve a Triangle Given Two Angles and a Side Answer: m A = 27°, a  5.1, and c  11.1 Example 2

Solve ΔABC. Round to the nearest tenth if necessary. A. B = 70°; a  4.7; b  17.0 B. B = 70°; a  3.3; b  11.1 C. B = 70°; a  5.2; b  15.3 D. B = 70°; a  9.7; b  12.5 Example 2

Concept

Solve a Triangle Given Two Sides and an Angle A. Determine whether ΔNPQ has no solution, one solution, or two solutions. Then solve the triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. In ΔNPQ, Q = 110°, q = 11, and n = 8. Example 3

Because Q is obtuse and 11 > 8, you know that one solution exists. Solve a Triangle Given Two Sides and an Angle Because Q is obtuse and 11 > 8, you know that one solution exists. Step 1 Use the Law of Sines to find mN. Law of Sines Multiply each side by 8. sin N ≈ 0.6834 Use a calculator. N ≈ 43 Use the sin–1 function. Example 3

Step 3 Use the Law of Sines to find p. Solve a Triangle Given Two Sides and an Angle Step 2 Find mP. mP = 180° – (110° + 43°) or 27° Step 3 Use the Law of Sines to find p. Law of Sines Solve for p. p ≈ 5.3 Use a calculator. Answer: So, One; N ≈ 43, P ≈ 27°, and p ≈ 5.3. Example 3

Solve a Triangle Given Two Sides and an Angle B. Determine whether ΔDEF has no solution, one solution, or two solutions. Then solve the triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. In ΔDEF, E = 52°, e = 5, and f = 9. Example 3

Since E is acute and 5 < 9, find h and compare it to e. Solve a Triangle Given Two Sides and an Angle Since E is acute and 5 < 9, find h and compare it to e. f sin E = 9 sin 52° f = 9 and E = 52° ≈ 7.1 Use a calculator. Answer: Since 5 < 7.1 or e < h, there is no solution. Example 3

Solve a Triangle Given Two Sides and an Angle C. Determine whether ΔXYZ has no solution, one solution, or two solutions. Then solve the triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. In ΔXYZ, X = 28°, z = 15, and x = 9. Example 3

Since X is acute and 9 < 15, find h and compare it to x. Solve a Triangle Given Two Sides and an Angle Since X is acute and 9 < 15, find h and compare it to x. f sin X = 15 sin 28° z = 15 and X = 28° ≈ 7.0 Use a calculator. Since 7.0 < 9 < 15 or h < x < z, there are two solutions. So, there are two triangles to be solved. Case 1 Z is acute. Example 3

Step 1 Find mZ. Law of Sines Solve for sin Z. Use a calculator. Solve a Triangle Given Two Sides and an Angle Step 1 Find mZ. Law of Sines Solve for sin Z. Use a calculator. Find sin–10.7825. Example 3

Step 2 Find mY. mY ≈ 180° – (28° + 51°) or 101° Step 3 Find y. Solve a Triangle Given Two Sides and an Angle Step 2 Find mY. mY ≈ 180° – (28° + 51°) or 101° Step 3 Find y. Law of Sines Solve for y. Simplify. Example 3

Case 2 Z is obtuse. Step 1 Find mZ. Solve a Triangle Given Two Sides and an Angle Case 2 Z is obtuse. Step 1 Find mZ. The sine function also has a positive value in Quadrant II. So, find an obtuse angle for B for which sin B ≈ 0.7825. mZ ≈ 180° – 51° or 129° Step 2 Find mY. mY ≈ 180° – (28° + 129°) or 23° Example 3

Step 3 Find y. Law of Sines Solve for y. Simplify. Solve a Triangle Given Two Sides and an Angle Step 3 Find y. Law of Sines Solve for y. Simplify. Answer: So, two; m Z ≈ 51°, m Y ≈ 101°, y ≈ 18.8; m Z ≈ 129°, m Y ≈ 23°, and y ≈ 7.5. Example 3

A. Determine whether ΔFGH has no solution, one solution, or two solutions. Then solve the triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. In ΔFGH, F = 36°, g = 9, and f = 6. A. no solution B. one solution; G ≈ 70°, H ≈ 74°, h = 9.8 C. one solution; G ≈ 110°, H ≈ 34°, h = 5.7 two solutions; G ≈ 62°, H ≈ 82°, h = 10.1; G ≈ 118°, H ≈ 26°, h = 4.5 Example 3

B. Determine whether ΔLMN has no solution, one solution, or two solutions. Then solve the triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. In ΔLMN, L = 100°, m = 7, and l = 10. A. no solution B. one solution; M ≈ 44°, N ≈ 36°, n = 6.0 C. one solution; M ≈ 49°, H ≈ 31°, n = 8.3 D. two solutions; M ≈ 44°, N ≈ 36°, n = 6.0; M ≈ 49°, N ≈ 31°, and n = 8.3 Example 3

C. Determine whether ΔWUV has no solution, one solution, or two solutions. Then solve the triangle. Round side lengths to the nearest tenth and angle measures to the nearest degree. In ΔWUV, W = 42°, u = 16, and w = 9. A. no solution B. one solution; U ≈ 51°, V ≈ 87°, v = 10.2 C. one solution; U ≈ 129°, V ≈ 9°, v = 8.1 D. two solutions; U ≈ 51°, V ≈ 87°, v = 10.2; U ≈ 129°, V ≈ 9°, and v = 8.1 Example 3

Use the Law of Sines to Solve a Problem BASEBALL A baseball is hit between second and third bases and is caught at point B, as shown in the figure. How far away from second base was the ball caught? Example 4

Answer: So, the distance is about 59.8 ft. Use the Law of Sines to Solve a Problem Law of Sines Cross products. Solve for x. Use a calculator. Answer: So, the distance is about 59.8 ft. Example 4

BASEBALL A baseball is hit between second and third bases and is caught at point B, as shown in the figure. How far away from third base was the ball caught? A. 69.3 ft B. 72.8 ft C. 76.7 ft D. 78.9 ft Example 4