Download Document

Five-Minute Check (over Lesson 8–5)
NGSSS
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
Name the angle of
depression in the figure.
A. URT
0%
B
A
D. SRU
0%
0%
A
B
C
D
0%
D
C. RST
A.
B.
C.
D.
C
B. SRT
Over Lesson 8–5
Find the angle of elevation of the Sun when a
6-meter flagpole casts a 17-meter shadow.
A. about 70.6°
0%
B
A
D. about 19.4°
0%
A
B
C
0%
D
D
C. about 29.6°
A.
B.
C.
0%
D.
C
B. about 60.4°
Over Lesson 8–5
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°
D. about 88.6°
0%
B
A
0%
A
B
C
0%
D
D
C. about 82.4°
C
B. about 2.4°
A.
B.
C.
0%
D.
Over Lesson 8–5
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
D. about 804 ft
0%
B
A
0%
A
B
C
0%
D
D
C. about 726 ft
C
B. about 236.4 ft
A.
B.
C.
0%
D.
Over Lesson 8–5
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
D. 4365 ft
0%
B
A
0%
A
B
C
0%
D
D
C. 1060 ft
C
B. 104 ft
A.
B.
C.
0%
D.
MA.912.T.2.1 Define and use the
trigonometric ratios (sine, cosine, tangent,
cotangent, secant, cosecant) in terms of
angles of right triangles.
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
Cross Products Property
Divide each side by sin
Use a calculator.
Answer: p ≈ 4.8
Find c to the nearest tenth.
A. 4.6
B. 29.9
0%
B
A
0%
A
B
C
0%
D
D
D. 8.5
C
C. 7.8
A.
B.
C.
0%
D.
Law of Sines (SSA)
Find x. Round to the nearest degree.
Law of Sines (SSA)
Law of Sines
mB = 50, b = 10, a = 11
Cross Products Property
Divide each side by 10.
Use the inverse sine ratio.
Use a calculator.
Answer: x ≈ 57.4
Find x. Round to the nearest degree.
A. 39
B. 43
0%
B
A
0%
A
B
C
0%
D
D
D. 49
C
C. 46
A.
B.
C.
0%
D.
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)
Law of Cosines
Simplify.
Take the square root
of each side.
Use a calculator.
Answer: x ≈ 18.9
Find r if s = 15, t = 32, and mR = 40. Round to the
nearest tenth.
A. 25.1
B. 44.5
A
B
C
0%
D
D
0%
B
0%
A
D. 21.1
C
C. 22.7
A.
B.
C.
0%
D.
Law of Cosines (SSS)
Find mL. Round to the nearest degree.
Law of Cosines
Simplify.
Law of Cosines (SSS)
Subtract 754 from each side.
Divide each side by –270.
Solve for L.
Use a calculator.
Answer: mL ≈ 49
Find mP. Round to the nearest degree.
A. 44°
B. 51°
0%
B
A
0%
A
B
C
0%
D
D
D. 69°
C
C. 56°
A.
B.
C.
0%
D.
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
Use the Law of Sines to find KJ.
Law of Sines
Cross products
Indirect Measurement
Divide each side by sin
.
Simplify.
Answer: The width of each wing is about 16.9 meters.
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.
A.
B.
C.
D.
B. 103.5 in.
C. 96.7 in.
D. 88.8 in.
0%
D
0%
C
0%
B
A
0%
A
B
C
D
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 mP.
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
–66 = –126 cos P
Simplify.
Subtract 130 from
each side.
Divide each side
by –126.
Use the inverse
cosine ratio.
Use a calculator.
Solve a Triangle
Use the Law of Sines to find mQ.
Law of Sines
mP ≈ 58, p = 8,
q = 7.
Multiply each side
by 7.
Use the inverse
sine ratio.
Use a calculator.
Solve a Triangle
By the Triangle Angle Sum Theorem,
mR ≈ 180 – (58 + 48) or 74.
Answer: Therefore, mP ≈ 58; mQ ≈ 48 and
mR ≈ 74.
Solve ΔRST. Round to the nearest degree.
A. mR = 82, mS = 58,
mT = 40
B. mR = 58, mS = 82,
mT = 40
A.
B.
C.
D.
0%
C
0%
B
D. mR = 40, mS = 58,
mT = 82
A
0%
0%
D
C. mR = 82, mS = 40,
mT = 58
A
B
C
D