Download then no triangle exists. - Fort Thomas Independent Schools

P.o.D. – Find the amplitude,
period, and vertical translation
of each graph.
1.) 𝑦 = 3 sin(2𝑥 ) − 5
2.) 𝑦 = cos(𝑥)
3.) 𝑦 = −4 sin(0.5𝑥 ) + 2
4.) 𝑦 = −2 cos(3𝑥 ) − 6
1
1
3
3
5.) 𝑦 = sin ( 𝑥) +
6.) 𝑦 = − cos(6𝑥 ) − 7
7.) 𝑦 = sin(𝑥 ) + 𝜋
1.)
2.)
3.)
4.)
A=3, P=180, VT=-5
A=1, P=360, VT=0
A=4, P=720, VT=2
A=2, P=120, VT=-6
5.) A=1, P=1080, VT=1/3
6.) A=1, P=60, VT=-7
7.) A=1, P=360, VT=pi
10-7: The Law of Sines
Learning Target(s): I can
determine the measure of an
angle given its sine, cosine, or
tangent; find the missing sides
and angles of a triangle using the
Law of Sines; identify and use
theorems relating sines and
cosines; solve real-world
problems using the Law of Sines.
Essential Question: How do you
use trigonometry to solve and
find the areas of oblique
triangles?
Oblique Triangle = a triangle
that has no right angle; could be
acute or obtuse.
Law of Sines:
C
b
A
a
c
B
𝑎
𝑏
𝑐
=
=
sin 𝐴 sin 𝐵 sin 𝐶
EX: For triangle ABC, A=30
degrees, B=45 degrees, and a=32
feet. Find the remaining sides
and angle.
Always begin by drawing a
picture.
B
45
32
30
A
C
The easiest thing to find is the
missing angle.
∠𝐶 = 180° − 30° − 45° = 105°
Now, use the given information
to find side b.
𝑏
32
=
→
sin 45° sin 30°
32 sin 45°
𝑏=
=
sin 30°
32√2 ≈ 45.25
Next, find side c.
𝑐
32
=
→
sin 105° sin 30°
32 sin 105°
𝑐=
=
sin 30°
61.82
Finally, write your solution
with all 6 parts of the triangle
together.
A=30 deg. B=45 deg. C=105
deg.
a=32
b=45.25
c=61.82
EX: Because of prevailing winds,
a tree grew so that it was
leaning 6 degrees from the
vertical. At a point 30 meters
from the tree, the angle of
elevation to the top of the tree is
22.5 degrees. Find the height of
the tree.
h
96
22.5
30
Begin by finding the missing
angle.
180-22.5-96=61.5 degrees
Set up the Law of Sine to find h.
ℎ
30
=
→ℎ
sin 22.5° sin 61.5°
30 sin 22.5°
=
=
sin 61.5°
13.06 𝑚𝑒𝑡𝑒𝑟𝑠
EX: For the triangle ABC,
a=12inches, b=5inches, and A=31
degrees. Find the remaining
sides and angles.
We need to find angle B.
sin 𝐵 sin 31
=
5
12
5 sin 31
→ sin 𝐵 =
→𝐵
12
5 sin 31
−1
= 𝑠𝑖𝑛 (
) = 𝑠𝑖𝑛−1 (.215)
12
= 12.39°
Next, find angle C by
subtraction.
C=180-12.39-31=136.61 degrees.
Now, find side c using the Law
of Sines.
𝑐
12
=
→𝑐
sin 136.61° sin 31°
12 sin 136.61°
=
= 16.01
sin 31°
Write your final answer.
A=31 deg. B=12.39
C=136.61
deg
deg.
a=12
b=5
c=16.01
The Ambiguous Case of the Law
of Sines (SSA):
Why couldn’t you prove triangle
congruency with SSA in
geometry?
(Draw a picture on the
whiteboard)
Three possibilities will exist: 1
solution (right triangle), 2
solutions (one acute, one
obtuse), or no triangle exists.
Case I: If A is an acute angle:
If 𝑎 < 𝑏 sin 𝐴, then no triangle
exists.
If 𝑎 = 𝑏 sin 𝐴, then exactly one
right triangle exists.
If 𝑎 > 𝑏 sin 𝐴, then two triangles
exist.
If 𝑎 ≥ 𝑏, then one triangle exists.
Case II: If A is an obtuse angle:
If a < b, then no triangle exists.
If a > b, then one triangle exists.
EX: Show that there is no
triangle for which A=60degrees,
a=4, and b=14.
We are in Case I since A is acute.
√3
𝑏 sin 𝐴 = 14 sin 60° = 14 ( ) = 7√3
2
≈ 12.124
In Case I, if a<b sin A, then we
have no solution. In this
example, 4<12.124.
EX: Find two triangles for
which A=58degrees, a=4.5, and
b=5.
Again, we are in Case I since A is
acute. We now need to find b sin
A.
𝑏 sin 𝐴 = 5 sin 58° ≈ 4.24
Since a>b sin A {4.5>4.24}, we
know that we have two
solutions. Begin by solving using
the Law of Sines.
Triangle #1:
sin 𝐵 sin 58°
5 sin 58°
=
→ sin 𝐵 =
5
4.5
4.5
5 sin 58°
−1
→ 𝐵 = 𝑠𝑖𝑛 (
)
4.5
≈ 70.437°
Use subtraction to find angle C.
𝐶 = 180 − 70.437 − 58 = 51.563°
Now, use the Law of Sines to
find side c.
𝑐
4.5
=
→𝑐
sin 51.563° sin 58°
4.5 sin 51.563°
=
≈ 4.156
sin 58°
Therefore, the first triangle
consists of:
A=58 deg. B=70.437
deg.
a=4.5
b=5
Triangle #2:
C=51.563
deg.
c=4.156
∠𝐵2 = 180 − 𝐵1 = 180 − 70.437
= 109.563°
In other words, the second angle
B is the supplement of the first
angle B.
Find angle 𝐶2 by subtraction.
𝐶2 = 180 − 109.563 − 58 = 12.437°
Lastly, find side 𝑐2 using the Law
of Sines.
𝑐2
4.5
=
→ 𝑐2
sin 12.437° sin 58°
4.5 sin 12.437°
=
≈
sin 58°
1.143
Triangle #2 consists of:
𝐴 = 58°
𝐵2
𝐶2
= 109.563° = 12.437°
𝑎 = 4.5
𝑏=5
𝑐2 = 1.143
Area of a Triangle:
The area of any triangle is onehalf the product of the lengths
of two sides times the sine of
their included angle. That is,
1
1
𝐴𝑟𝑒𝑎 = 𝑎𝑏 sin 𝐶 = 𝑏𝑐 sin 𝐴
2
2
1
= 𝑎𝑐 sin 𝐵
2
EX: Find the area of a triangular
lot containing the side lengths
that measure 24 yards and 18
yards and form an angle of 80
degrees.
1
𝐴 = (24)(18) sin 80°
2
≈ 212.718 𝑠𝑞𝑢𝑎𝑟𝑒 𝑦𝑎𝑟𝑑𝑠.
*On your own, write a
calculator program titled
SASAREA that will compute the
area of a triangle given two sides
and the included angle.
EX: On a small lake, a child
swam from point A to point B at
a bearing of 𝑁 28° 𝐸. The child
then swam to point C at a
bearing of 𝑁 58° 𝑊. Point C is 800
meters due north of point A.
How many total meters did the
child swim?
(draw a picture and show the
work on the whiteboard)
About 1056 meters.
EX: Suppose you are given a
triangle with two angles of 42
degrees and 74 degrees and an
included side of 18-m. Find the
length of the side opposite the
42 degrees.
(draw a triangle on the
whiteboard and have a student
come to the board to solve)
About 13.4m
EX: In triangle TRI, angle R is 25
degrees, angle I is 72 degrees,
and side r is 13. Find side t.
(draw a triangle on the board
and have a student come to the
board to solve)
t=30.53
Upon completion of this lesson,
you should be able to:
1. Solve triangles using the
Law of Sines.
2. Find the area of a triangle
given two side and the
included angle.
For more information, visit
http://www.mathsisfun.com/algebra/trigsine-law.html
HW Pg.703 2, 7-12, 15-18
Worksheet 10.7B