Transcript Slide 1

(8 – 2)
Law of Sines
Learning target: To solve SAA or ASA triangles
To solve SSA triangles
To solve applied problems
We work with any triangles:
C
Important how to label.
We use capital letters for angles
(vertexes) and small letters for
sides corresponding.
a
b
A
c
B
Opposite side of A is a.
Opposite side of B is b.
Opposite side of C is c.
Cases to use the law of sines:
Case 1: (AAS or ASA) two angles and one side are known.
• If you are given two angles and one side (ASA or AAS),
the Law of Sines will nicely provide you with ONE solution
for a missing side.
Case 2: (SSA) two sides and an angle opposite one of them are
known.
• Unfortunately, the Law of Sines has a problem dealing with
SSA.
If you are given two sides and one angle (where you
must find an angle), the Law of Sines could possibly provide you
with one or more solutions, or even no solution.
Case 1: AAS case
I do: Solve for all missing parts.
1: label all missing parts
2: find the missing angle.
c
60
4
40
b
3: set up the equation using law of sine,
and solve for missing parts.
ASA case
We do: Solve for all missing parts.
1: label all missing parts.
C
10
45
A
2: find the remaining angle.
105
3: set up the equation using
B law of sine, and solve for
missing parts.
You do: Solve for all missing parts.
29
C
1:leble all missing parts
2: find the case.
43 cm
3: find the remaining angle
52
B
A
4:set up the equation using law of
sine, and solve for missing parts.
You do: Solve for all missing parts.
1:leble all missing parts
A
2: find the case.
3: find the remaining angle
94.6 m
18.7
4:set up the equation using law of B
sine, and solve for missing parts.
124.1
C
Case 2: (SSA) two sides and an angle opposite one of them are
known.
Unfortunately, the Law of Sines has a problem dealing with
SSA.
If you are given two sides and one angle (where you
must find an angle), the Law of Sines could possibly provide you
with one or more solutions, or even no solution.
In Geometry, we found that we could prove two triangles congruent using:
SAS - Side, Angle, Side
ASA - Angle, Side, Angle
AAS - Angle, Angle, Side
SSS - Side, Side, Side
HL - Hypotenuse Leg for
Right Triangles.
We also discovered that
SSA did not work to prove
triangles congruent.
We politely called it the
Donkey Theorem ; - )
a<h
y
b
C
h = b sinA
a=h
h
a
a (h)
A
no triangle is constructed
The arc intersects at 2distinct
positive-x points.
h<a<b
x
B
A right triangle is constructed
The arc intersects at 1 positive-x
point.
a>b
a
a
a h
B
B
2 triangles are constructed
B
One non-right triangle is constructed
Facts we need to remember:
1.
I do: Find all missing parts. (SSA)
1: label all parts.
C
12.4
2: find angle A using law of
sine.
8.7
36.7
B
A
3: find A.
5: write the 2 triangles.
ABC & A’BC’
6: find the remaining parts for the 2
triangles.
4: check if there is another
possible angle.
We have 2 triangles.
C’
6: find the remaining parts for the
2 triangles.
ABC:
• find C :
C  84.9
• find c using law of sine
6: A’BC’:
•
find C’ :
C’  21.7
• Find c’ using law os sine
We do: Find all missing parts. (SSA)
1: label all parts.
C
8
2: find angle B using law of
sine.
6
6
35
A
B’
B
3: find B.
5: draw and label, and write the 2
triangles.
ABC & A’BC’
6: find the remaining parts for the 2
triangles.
B 
4: check if there is another
possible angle.
Let
We have 2 triangles.
C’
6: find the remaining parts for the
2 triangles.
ABC:
c’
• find C using the sum of a
triangle = 180
6: A’BC’:
•
find B’ :
• Find c’ using law os sine
find c using law of sine
You do: Find the unknown parts.
A = 29.7, b = 41.5 ft, a = 27.2 ft.
<Try this>
Find all missing parts.
184.5 ft
60
123 ft
<try this>
P
21
A
65
B
A cable car carries passengers
from a point A to point P. The
point A is 1.2 miles from a
point B at the base of a
mountain. The angles of
elevation of P from A and B
are 21 and 65, respectively.
a) Approximate the distance between A and P.
b) Approximate the height of the mountain.
(8 – 3)
Law of cosines
Learning target: To solve SAS triangles
To solve SSS triangles
To solve applied problems.
Cases to use law of cosine
Case 1: (SAS)
two sides and the angle between them are known.
Case 2: (SSS)
three sides are known.
This may help you memorize.
There is a pattern in the formulas.
• all letters of sides, squared letters are multiplied with 2cosine.
• outside letters are the same
Law of cosine, case 1: (SAS case)
I do: Solve the triangle if a = 5, c = 8, and B = 77
B
5
77
C
8
1: draw an triangle and label given
number for each part.
2: solve for b using law of cosine. (since
B = 77 is given)
A
4: find C using the sum
of .
b  8.4
3: Find A using law of sine.
Law of cosine, case 2: (SSS case)
We do: Approximate the angles of the triangle if a = 90, b = 70,
and c = 40
1: draw an triangle and label given
number for each part.
2: find one angle using law of cosine.
3: Find another side using law of cosine
again.
4: find the last side using the sum of .
You do: In a  side b = 12, side c = 20 and m A = 45º. Solve the
triangle.
1: draw an triangle, and label given number for each part.
2: find the case (SAS or SSS)
3: find the missing side.
4: find other missing angles.
<try this>
a) Find the largest angle, to the nearest tenth of a degree, of a triangle
whose sides are 9, 12 and 18.
b) In a parallelogram, the adjacent sides measure 40 cm and 22 cm. If
the larger angle of the parallelogram measure 116º, find the length of
the larger diagonal, to the nearest integer.
(8 – 4)
Area of a Triangle
Learning target: To find the area of SAS triangle
To find the area of SSS triangles
Heron’s Formula for the area of a triangle:
If the three sides of a triangle are a, b, and
c, then the area of the triangle is:
K  s(s  a)(s  b)(s  c)
where
1
s  (a  b  c)
2
a
c
b
Cases to use Heron’s formula:
Case 1: (SAS)
two sides and the angle between them are known.
Case 2: (SSS)
three sides are known.
I do: Find the area of the triangle for which a = 8, b = 6, and C=30
B
1: draw a triangle, and label all parts.
8
2: find c using law of cosine.
c
30
C
3: find the area K using Heron’s formula.
6
A
We do: Find the area of the triangle for which a = 4, b = 5, and c = 7.
1: write the Heron’s formula.
K  s(s  a)(s  b)(s  c)
2: Find the perimeter, and find s.
1
s  (a  b  c) = 8
2
3: plug in the value of s, and find the area.
You do: find the area of the triangle.
a) Given a = 154 cm, b = 179 cm, c = 183 cm.
<try this> Solve.
a) A painter needs to cover a triangular region 75 m by 68 m by
85
m. A can of paint covers 75 sq m of area. How many cans will be
needed?
b) Find the area of a triangle in a rectangular coordinate plane
whose vertices are (0, 0), (3, 4), and (-8, 6) using Heron’s formula.
Hint: draw a triangle, and write the coordinates, then find the
each side using the distance formula.