Transcript Engineering math Review • Trigonometry • Systems of Equations • Vectors • Vector Addition and Subtraction • Vector Multiplication.

Engineering math
Review
• Trigonometry
• Systems of Equations
• Vectors
• Vector Addition and Subtraction
• Vector Multiplication
Trigonometry
Review
• Pythagorean Triples
• Sine, Cosine, and Tangent revisited.
• Law of Sines, Law of Cosines
Starting With Right
Triangles
c
a
b
Starting With Right
Triangles
c
a
b
Recall the Pythagorean Theorem
a
2
+b =c
2
2
Starting With Right
Triangles
c
a
b
a +b = c
2
2
Pythagorean Triples
There are a few sets of numbers that satisfy the Pythagorean
Theorem using only whole numbers for a, b, and c.
c
a
b
a
3
b
4
c
5
5
8
12
15
13
17
Sine
c
a
a
b
opposite
a
sin(a ) =
=
hypotenuse c
Cosine
c
a
a
b
adjacent
b
cos(a ) =
=
hypotenuse c
tangent
c
a
a
b
opposite a
tan(a ) =
=
adjacent b
Inverse Trig
Functions
Recall that inverse trig functions are used to solve for the angle, α, after setting
up the trig equations
c
a
α
b
For example:
a
sin(a ) =
c
so then
æaö
a = sin ç ÷
ècø
-1
Law of Sines
B
c
a
A
C
b
sin ( A) sin ( B) sin (C )
=
=
a
b
c
Because of the properties of the sine function between 0 and 180, the Law of
Sines may yield 2 answers. If possible, its best to use the Law of Sines when
you know whether an angle is acute (less than 90) or obtuse (greater than 90)
Law of Cosines
B
The Law of Cosines can be used
for obtuse angles (angles greater
than 90 degrees) as well as acute
angles
c
a
A
C
b
a = b + c - 2bccos(A)
2
2
Law of Cosines
B
The Law of Cosines can be used
for obtuse angles (angles greater
than 90 degrees) as well as acute
angles
c
a
A
C
b
b = a + c - 2accos(B)
2
2
Law of Cosines
B
The Law of Cosines can be used
for obtuse angles (angles greater
than 90 degrees) as well as acute
angles
c
a
A
C
b
c = a + b - 2abcos(C)
2
2
Example Problem
B
Solve for all angles of the
triangle pictured to the left
c = 12
a=9
A
C
b = 18
Example Problem
B
Solve for all angles of the
triangle pictured to the left
c = 12
a=9
A
C
b = 18
b = a + c - 2accos(B)
2
2
Start by using Law of Cosines to
determine angle B. This will be the
largest angle in the triangle because it
is opposite to the longest side.
Example Problem
B
Solve for all angles of the
triangle pictured to the left
c = 12
a=9
A
C
b = 18
b = a + c - 2accos(B)
2
2
b = a + c - 2accos(B)
2
2
2
Our goal is to calculate angle B. To
do this, we’ll need to isolate the
expression with angle B in it.
We start by squaring both sides of the
equation
Example Problem
B
Solve for all angles of the
triangle pictured to the left
c = 12
a=9
Subtract the a2 and c2 terms next.
A
C
b = 18
b = a + c - 2accos(B)
2
2
b = a + c - 2accos(B)
b2 - a2 - c 2 = -2accos(B)
2
2
2
Example Problem
B
Solve for all angles of the
triangle pictured to the left
c = 12
a=9
Divide both sides by -2ac
A
C
b = 18
b = a + c - 2accos(B)
2
2
b = a + c - 2accos(B)
b2 - a2 - c 2 = -2accos(B)
2
2
2
b2 - a2 - c2
= cos(B)
-2ac
Example Problem
B
Solve for all angles of the
triangle pictured to the left
c = 12
a=9
A
C
b = 18
b = a + c - 2accos(B)
2
2
b = a + c - 2accos(B)
b2 - a2 - c 2 = -2accos(B)
2
Take the inverse cosine of both sides
of the equation to isolate the angle of
interest, angle B.
2
2
b2 - a2 - c2
= cos(B)
-2ac
2
2
2ö
æ
b
a
c
cos-1 ç
÷= B
è -2ac ø
Example Problem
B
Solve for all angles of the
triangle pictured to the left
c = 12
a=9
A
C
b = 18
b2 - a2 - c2
= cos(B)
-2ac
2
2
2ö
æ
b
a
c
-1
cos ç
÷= B
è -2ac ø
Substitute the known lengths of the
sides of the triangle and solve for
angle B
æ 182 - 9 2 -12 2 ö
÷÷ = B
cos-1 çç
è -2 ( 9) (12) ø
Example Problem
B
Solve for all angles of the
triangle pictured to the left
c = 12
a=9
A
C
b = 18
b2 - a2 - c2
= cos(B)
-2ac
2
2
2ö
æ
b
a
c
-1
cos ç
÷= B
è -2ac ø
Substitute the known lengths of the
sides of the triangle and solve for
angle B
æ 182 - 9 2 -12 2 ö
÷÷ = B
cos-1 çç
è -2 ( 9) (12) ø
-1 æ 99 ö
cos ç
÷= B
è -216 ø
Example Problem
B
Solve for all angles of the
triangle pictured to the left
c = 12
a=9
A
C
b = 18
b2 - a2 - c2
= cos(B)
-2ac
2
2
2ö
æ
b
a
c
-1
cos ç
÷= B
è -2ac ø
Substitute the known lengths of the
sides of the triangle and solve for
angle B
æ 182 - 9 2 -12 2 ö
÷÷ = B
cos-1 çç
è -2 ( 9) (12) ø
-1 æ 99 ö
cos ç
÷= B
è -216 ø
117.3° = B
Example Problem
B=
117.3°
a=9
Solve for all angles of the
triangle pictured to the left
c = 12
A
C
b = 18
sin ( A) sin ( B)
=
a
b
Now that we know one of the angles,
and we know that the remaining
angles have to be less than 90°, we
can use Law of Sines to determine
angle A
Example Problem
B=
117.3°
a=9
Solve for all angles of the
triangle pictured to the left
c = 12
A
C
b = 18
sin ( A) sin ( B)
=
a
b
éa
ù
sin ( A) = ê sin(B)ú
ëb
û
Multiplying both sides by ‘a’ isolates
the term with angle A in it
Example Problem
B=
117.3°
a=9
Solve for all angles of the
triangle pictured to the left
c = 12
A
C
b = 18
sin ( A) sin ( B)
=
a
b
éa
ù
sin ( A) = ê sin(B)ú
ëb
û
Taking the inverse sine of both sides
of the equation solves in terms of
angle A
éa
ù
A = sin ê sin(B)ú
ëb
û
-1
Example Problem
B=
117.3°
a=9
Solve for all angles of the
triangle pictured to the left
c = 12
A
C
b = 18
sin ( A) sin ( B)
=
a
b
éa
ù
sin ( A) = ê sin(B)ú
ëb
û
Substitute known values to solve for
angle A
éa
ù
A = sin ê sin(B)ú
ëb
û
ù
-1 é 9
A = sin ê sin(117.3°)ú
ë18
û
-1
Example Problem
B=
117.3°
a=9
Solve for all angles of the
triangle pictured to the left
c = 12
A
C
b = 18
sin ( A) sin ( B)
=
a
b
éa
ù
sin ( A) = ê sin(B)ú
ëb
û
Substitute known values to solve for
angle A
éa
ù
A = sin ê sin(B)ú
ëb
û
ù
-1 é 9
A = sin ê sin(117.3°)ú
ë18
û
-1
A = 26.4°
Example Problem
B=
117.3°
a=9
c = 12
A = 26.4°
C
b = 18
sin (C ) sin ( B)
=
c
b
Solve for all angles of the
triangle pictured to the left
We can use Law of Sines to
determine angle C or capitalize on
the fact that A + B + C = 180°. We’ll
use Law of Sines, and then check the
results using A + B + C = 180°
Example Problem
B=
117.3°
a=9
c = 12
A = 26.4°
C
b = 18
sin (C ) sin ( B)
=
c
b
éc
ù
sin (C ) = ê sin(B)ú
ëb
û
Solve for all angles of the
triangle pictured to the left
Multiplying both sides by ‘c’ isolates
the term with angle C in it
Example Problem
B=
117.3°
a=9
c = 12
A = 26.4°
C
b = 18
sin (C ) sin ( B)
=
c
b
éc
ù
sin (C ) = ê sin(B)ú
ëb
û
Solve for all angles of the
triangle pictured to the left
Taking the inverse sine of both sides
of the equation solves in terms of
angle C
éc
ù
C = sin ê sin(B)ú
ëb
û
-1
Example Problem
B=
117.3°
a=9
c = 12
A = 26.4°
C
b = 18
sin (C ) sin ( B)
=
c
b
éc
ù
sin (C ) = ê sin(B)ú
ëb
û
Solve for all angles of the
triangle pictured to the left
Substitute known values to solve for
angle C
éc
ù
C = sin ê sin(B)ú
ëb
û
ù
-1 é12
C = sin ê sin(117.3°)ú
ë18
û
-1
Example Problem
B=
117.3°
a=9
c = 12
A = 26.4°
C
b = 18
sin (C ) sin ( B)
=
c
b
éc
ù
sin (C ) = ê sin(B)ú
ëb
û
Solve for all angles of the
triangle pictured to the left
Substitute known values to solve for
angle C
éc
ù
C = sin ê sin(B)ú
ëb
û
ù
-1 é12
C = sin ê sin(117.3°)ú
ë18
û
-1
C = 36.3°
Example Problem
a=9
B=
117.3°
c = 12
A = 26.4°
C = 36.3°
Solve for all angles of the
triangle pictured to the left
Now, we can check our answer and
see if A + B + C = 180°
b = 18
A + B +C =180°
Example Problem
a=9
B=
117.3°
c = 12
A = 26.4°
C = 36.3°
Solve for all angles of the
triangle pictured to the left
Now, we can check our answer and
see if A + B + C = 180°
b = 18
A + B +C =180°
117.3+ 26.4 +36.3 =180°
Example Problem
a=9
B=
117.3°
c = 12
Solve for all angles of the
triangle pictured to the left
Everything checks out!
A = 26.4°
C = 36.3°
b = 18
A + B +C =180°
117.3+ 26.4 +36.3 =180°
180° =180°