350 likes | 375 Views
LESSON 8–6. The Law of Sines and the Law of Cosines. Five-Minute Check (over Lesson 8–5) TEKS Then/Now New Vocabulary Theorem 8.10: Law of Sines Example 1: Law of Sines (AAS) Example 2: Law of Sines (ASA) Theorem 8.11: Law of Cosines Example 3: Law of Cosines (SAS)
E N D
LESSON 8–6 The Law of Sines and the Law of Cosines
Five-Minute Check (over Lesson 8–5) TEKS Then/Now New Vocabulary Theorem 8.10: Law of Sines Example 1: Law of Sines (AAS) Example 2: Law of Sines (ASA) Theorem 8.11: Law of Cosines Example 3: Law of Cosines (SAS) Example 4: Law of Cosines (SSS) Example 5: Real-World Example: Indirect Measurement Example 6: Solve a Triangle Concept Summary: Solving a Triangle Lesson Menu
Name the angle of depression in the figure. A. URT B. SRT C. RST D. SRU 5-Minute Check 1
Find the angle of elevation of the Sun when a 6-meter flagpole casts a 17-meter shadow. A. about 70.6° B. about 60.4° C. about 29.6° D. about 19.4° 5-Minute Check 2
After flying at an altitude of 575 meters, a helicopter starts to descend when its ground distance from the landing pad is 13.5 kilometers. What is the angle of depression for this part of the flight? A. about 1.8° B. about 2.4° C. about 82.4° D. about 88.6° 5-Minute Check 3
The top of a signal tower is 250 feet above sea level. The angle of depression from the top of the tower to a passing ship is 19°. How far is the foot of the tower from the ship? A. about 81.4 ft B. about 236.4 ft C. about 726 ft D. about 804 ft 5-Minute Check 4
Jay is standing 50 feet away from the Eiffel Tower and measures the angle of elevation to the top of the tower as 87.3°. Approximately how tall is the Eiffel Tower? A. 50 ft B. 104 ft C. 1060 ft D. 4365 ft 5-Minute Check 5
Targeted TEKS G.9(A) Determine the lengths of sides and measures of angles in a right triangle by applying the trigonometric ratios sine, cosine, and tangent to solve problems. Mathematical Processes G.1(E), G.1(F) TEKS
You used trigonometric ratios to solve right triangles. • Use the Law of Sines to solve triangles. • Use the Law of Cosines to solve triangles. Then/Now
Law of Sines • Law of Cosines Vocabulary
Law of Sines (AAS) Find p. Round to the nearest tenth. We are given measures of two angles and a nonincluded side, so use the Law of Sines to write a proportion. Example 1
Divide each side by sin Law of Sines (AAS) Law of Sines Cross Products Property Use a calculator. Answer:p≈ 4.8 Example 1
Find c to the nearest tenth. A. 4.6 B. 29.9 C. 7.8 D. 8.5 Example 1
6 57° x Law of Sines (ASA) Find x. Round to the nearest tenth. Example 2
Law of Sines (ASA) Law of Sines mB = 50, mC = 73, c = 6 6 sin 50 = x sin 73 Cross Products Property Divide each side by sin 73. 4.8 = x Use a calculator. Answer:x ≈ 4.8 Example 2
x 43° Find x. Round to the nearest degree. A. 8 B. 10 C. 12 D. 14 Example 2
Law of Cosines (SAS) Find x. Round to the nearest tenth. Use the Law of Cosines since the measures of two sides and the included angle are known. Example 3
Law of Cosines (SAS) Law of Cosines Simplify. Take the square root of each side. Use a calculator. Answer:x ≈ 18.9 Example 3
Find r if s = 15, t = 32, and mR = 40. Round to the nearest tenth. A. 25.1 B. 44.5 C. 22.7 D. 21.1 Example 3
Law of Cosines (SSS) Find mL. Round to the nearest degree. Law of Cosines Simplify. Example 4
Law of Cosines (SSS) Subtract 754 from each side. Divide each side by –270. Solve for L. Use a calculator. Answer:mL ≈ 49 Example 4
Find mP. Round to the nearest degree. A. 44° B. 51° C. 56° D. 69° Example 4
Indirect Measurement AIRCRAFT From the diagram of the plane shown, determine the approximate width of each wing. Round to the nearest tenth meter. Example 5
Indirect Measurement Use the Law of Sines to find KJ. Law of Sines Cross products Example 5
Divide each side by sin . Indirect Measurement Simplify. Answer: The width of each wing is about 16.9 meters. Example 5
The rear side window of a station wagon has the shape shown in the figure. Find the perimeter of the window if the length of DB is 31 inches. Round to the nearest tenth. A. 93.5 in. B. 103.5 in. C. 96.7 in. D. 88.8 in. Example 5
Solve a Triangle Solve triangle PQR. Round to the nearest degree. Since the measures of three sides are given (SSS), use the Law of Cosines to find mP. p2 = r2 + q2 – 2pq cos P Law of Cosines 82 = 92 + 72 – 2(9)(7) cos P p = 8, r = 9, and q = 7 Example 6
Solve a Triangle 64= 130 – 126 cos P Simplify. –66= –126 cos P Subtract 130 from each side. Divide each side by –126. Use the inverse cosine ratio. Use a calculator. Example 6
Solve a Triangle Use the Law of Sines to find mQ. Law of Sines mP≈ 58, p = 8,q = 7 Multiply each side by 7. Use the inverse sine ratio. Use a calculator. Example 6
Solve a Triangle By the Triangle Angle Sum Theorem, mR≈ 180 – (58 + 48) or 74. Answer: Therefore, mP≈ 58; mQ≈ 48 andmR≈ 74. Example 6
Solve ΔRST. Round to the nearest degree. A.mR = 82, mS = 58, mT = 40 B.mR = 58, mS = 82, mT = 40 C.mR = 82, mS = 40, mT = 58 D.mR = 40, mS = 58, mT = 82 Example 6
LESSON 8–6 The Law of Sines and the Law of Cosines