splash screen. lesson menu five-minute check (over lesson 8–5) then/now new vocabulary theorem...
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Five-Minute Check (over Lesson 8–5)
Then/Now
New Vocabulary
Theorem 8.10: Law of Sines
Example 1: Law of Sines (AAS or ASA)
Example 2: Law of Sines (SSA)
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
Over Lesson 8–5
A. URT
B. SRT
C. RST
D. SRU
Name the angle of depression in the figure.
Over Lesson 8–5
A. about 70.6°
B. about 60.4°
C. about 29.6°
D. about 19.4°
Find the angle of elevation of the Sun when a 6-meter flagpole casts a 17-meter shadow.
Over Lesson 8–5
A. about 1.8°
B. about 2.4°
C. about 82.4°
D. about 88.6°
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?
Over Lesson 8–5
A. about 81.4 ft
B. about 236.4 ft
C. about 726 ft
D. about 804 ft
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?
Over Lesson 8–5
A. 50 ft
B. 104 ft
C. 1060 ft
D. 4365 ft
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?
You used trigonometric ratios to solve right triangles. (Lesson 8–4)
• Use the Law of Sines to solve triangles.
• Use the Law of Cosines to solve triangles.
• Law of Sines
• Law of Cosines
Law of Sines (AAS or ASA)
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.
Law of Sines (AAS or ASA)
Law of Sines
Use a calculator.
Divide each side by sin
Cross Products Property
Answer: p ≈ 4.8
A. 4.6
B. 29.9
C. 7.8
D. 8.5
Find c to the nearest tenth.
Law of Sines (SSA)
Find x. Round to the nearest degree.
Law of Sines (SSA)
Law of Sines
mB = 50, b = 10, a = 11
Cross Products Property
Divide each side by 10.
Use a calculator.
Use the inverse sine ratio.
Answer: x ≈ 57.4
A. 39
B. 43
C. 46
D. 49
Find x. Round to the nearest degree.
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.
Law of Cosines (SAS)
Answer: x ≈ 18.9
Simplify.
Take the square root of each side.
Law of Cosines
Use a calculator.
A. 25.1
B. 44.5
C. 22.7
D. 21.1
Find r if s = 15, t = 32, and mR = 40. Round to the nearest tenth.
Law of Cosines (SSS)
Find mL. Round to the nearest degree.
Law of Cosines
Simplify.
Law of Cosines (SSS)
Answer: mL ≈ 49
Solve for L.
Use a calculator.
Subtract 754 from each side.
Divide each side by –270.
A. 44°
B. 51°
C. 56°
D. 69°
Find mP. Round to the nearest degree.
Indirect Measurement
AIRCRAFT From the diagram of the plane shown, determine the approximate width of each wing. Round to the nearest tenth meter.
Indirect Measurement
Cross products
Law of Sines
Use the Law of Sines to find KJ.
Indirect Measurement
Answer: The width of each wing is about 16.9 meters.
Simplify.
Divide each side by sin .
A. 93.5 in.
B. 103.5 in.
C. 96.7 in.
D. 88.8 in.
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.
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
Solve a Triangle
64 = 130 – 126 cos P Simplify.
–66 = –126 cos P Subtract 130 fromeach side.
Divide each side by –126.
Use the inversecosine ratio.
Use a calculator.
Solve a Triangle
Use the Law of Sines to find mQ.
Law of Sines
Multiply each side by 7.
Use the inversesine ratio.
Use a calculator.
mP ≈ 58, p = 8,q = 7
Solve a Triangle
Answer: Therefore, mP ≈ 58; mQ ≈ 48 andmR ≈ 74.
By the Triangle Angle Sum Theorem, mR ≈ 180 – (58 + 48) or 74.
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
Solve ΔRST. Round to the nearest degree.