Transcript Law of Sines

Warm UP!
Solve for all missing angles and sides:
Z
5
3
Y
x
What formulas did you use to
solve the triangle?
• Pythagorean theorem
• SOHCAHTOA
• All angles add up to 180o in a triangle
Could you use those formulas on this
triangle?
Solve for all missing angles and sides:
This is an oblique triangle.
An oblique triangle is any
non-rightztriangle.
5
3
y
35o
x
There are formulas to solve oblique triangles just like
there are for right triangles!
LG 5-2 Solving Oblique
Triangles
Laws of Sines and Cosines
& Triangle Area
MM4A6. Students will solve trigonometric equations both
graphically and algebraically.
d. Apply the law of sines and the law of cosines.
LG 5-2 Solving Oblique Triangles
1. Laws of Sines and Cosines
2. Oblique Triangle Area
3. Practice 5-2/Review for Test
4. LG 5-2 TEST (1/31 ODD & 2/1 EVEN)
General Comments
C
You have learned to solve right
triangles in ACC Math 2.
Now we will solve oblique
triangles (non-right triangles).
Note: Angles are Capital letters
and the side opposite is the
same letter in lower case.
a
b
A
B
c
C
a
b
A
c
B
What we already know
• The interior angles total 180.
• We can’t use the Pythagorean
Theorem. Why not?
• For later, area = ½ bh
• Larger angles are across from
longer sides and vice versa.
• The sum of two smaller sides
must be greater than the third.
C
a
b
A
c
B
The Law of Sines helps you solve for sides or
angles in an oblique triangle.
sin A sin B sin C


a
b
c
(You can also use it upside-down)
a
b
c


sin A sin B sin C
Use Law of SINES when ...
…you have 3 parts of a triangle and you need to find the other
3 parts.
They cannot be just ANY 3 dimensions though, or you won’t
have enough info to solve the Law of Sines equation.
Use the Law of Sines if you are given:
• AAS - 2 angles and 1 adjacent side
• ASA - 2 angles and their included side
• ASS – (SOMETIMES) 2 sides and their adjacent angle
General Process for Law Of Sines
1. Except for the ASA triangle, you will
always have enough information for 1 full
fraction and half of another. Start with that
to find a fourth piece of data.
2. Once you know 2 angles, you can subtract
from 180 to find the 3rd.
3. To avoid rounding error, use given data
instead of computed data whenever
possible.
Example 1
Solve this triangle:
The angles in a ∆ total 180°,
so solve for angle C.
B
80°
c
A
Set up the Law of Sines to
find side b:
12
70°
b
12
b

sin 70 sin 80
12 sin 80  b sin 70
12 sin 80
b
 12.6cm
sin 70
C
12
c

sin 70 sin 30
12 sin 30  c  sin 70
12 sin 30
c
 6.4cm
sin 70
Angle C = 30°
Side b = 12.6 cm
Side c = 6.4 cm
Example 2: Solve this triangle
C
85
b
a =30
50
45
A
You’re given both pieces for sinA/a
and part of sinB/b, so we start
there. sin 45 sin 50
c
sin A sin C

a
c
sin 45 sin 85

30
c
c sin 45  30sin85
30sin85
c
sin 45

B
30
b
b sin 45  30sin 50
30sin 50
b
sin 45
Using a calculator, b  32.5
Using a calculator c  42.3
Example 3: Solve this triangle
C
Since we can’t start one of the
fractions, we’ll start by finding C.
11.1
C = 180 – 35 – 10 = 135
Since the angles were exact, this
isn’t a rounded value. We use
sinC/c as our starting fraction.
sin C sin A
sin C sin B

and

c
a
c
b
b
135
a
35
A
sin135 sin 35

45
a
a sin135  45sin 35
45sin 35
a
sin135
10
c
45
B
sin135 sin10

45
b
b sin135  45sin10
45sin10
b
sin135
Using your calculator
a  36.5
36.5
b  11.1
You try! Solve this triangle
B
30°
c
a = 30
C
115°
b
A
The Law of Cosines
When solving an oblique triangle, using one of
three available equations utilizing the cosine of
an angle is handy. The equations are as
follows:
a  b  c  2bc cos(A)
2
2
2
b  a  c  2ac cos(B)
2
2
2
c  a  b  2ab cos(C)
2
2
2
General Strategies for Using
the Law of Cosines
The formula for the Law of Cosines makes use
of three sides and the angle opposite one of
those sides. We can use the Law of Cosines:
• SAS -
two sides and the included angle
• SSS - all three sides
Example 1: Solve this triangle
87.0°
17.0
15.0
B
A
Now, since we know the
measure of one angle and
the length of the side
opposite it, we can use
the Law of Sines.
c
sin 87.0 sin A

22.1
15.0
Use the relationship:
c2 = a2 + b2 – 2ab cos C
c2 = 152 + 172 – 2(15)(17)cos(87°)
c2 = 487.309…
c = 22.1
  42.7
sin 87.0 sin B

22.1
17.0
  50.2
Example 2: Solve this triangle
C
sin B sin 36.9

31.4
23.2
23.2
31.4


38.6
We start by finding cos A.
a 2  b 2  c 2  2bc cos A
sin C sin 36.9

38.6
23.2
cos A  0.7993
A  36.9
  54.4
C  87.3
You TRY:
1. Solve a triangle with a = 8, b =10, and c = 12.
A = 41.4o
B = 55.8o
C = 82.8o
a= 8
b = 10
c = 12
2. Solve a triangle with A = 88o, B =16o, and c = 14.
A = 88o
B = 16o
C = 76o
a = 12.4
b = 3.4
c = 14
IMPORTANT
• IT IS ALWAYS BEST TO USE LAW OF SINES
FOR SIDES AND LAW OF COSINES FOR
ANGLES
• Sometimes, however, it is just not possible – you
may have to switch it up
Practice
• Do multiples of 3 in class.
• Turn in your answer to #6 and #15
• Complete the rest for HW