Download Triangle Summary

Similar Triangles
B
A
b
a
Compare Like Ratios to Like Ratios
c
Examples:
Large A B
Left
=
or
Small a
b
Right
C
corresponding angles
match
Slope
Grade
Pitch
Gradient
Tangent
A a
= etc.
B b
imbedded triangles are
similar when P || Q
m=
rise
m=
run
45°-45°-90° Triangles
y
x
reflected triangles are
similar when P || Q
x2 + y2 = z2
z
1 + m2
x=
a + b = 90°
30°-60°-90° Triangles
Heron's Formula
b
a
zm
1 + m2
y=
A
s=
c
A=
a
c
n=sides
Q
Δy = y2 − y1
d
B
Δz = z2 − z1
total degrees = (n–2) 180°
D2 = Δx2 + Δy2+ Δz2
A = π R2
a
As
rc
R
A
Δx = x2 − x1
e
C
s(s–a)(s–b)(s–c)
P
3D Pythagorean Theorem
b
a + b +c
2
D=2R
D
C=πD
C=2πR
R
θ
π R2
360°
θ
arc =
2πR
360°
AS =
B1
B2
+
(B1 + B2)
A=
H
2
A=
|[
slope = m =
D=
b
B
]
–
2
y1
y2
y3
y4
y1
Parallelograms
x1 y2
x2 y3
x3 y4
x4 y5
[
]
+
|
A=BH
a + b = 180°
x1 + x2 y1 + y2
,
)
2
2
Δx2 + Δy2
b
B
4 π R3
3
rise Δy y2 – y1
=
=
run
Δx x2 – x1
midpoint = (
Volume of
Frustums
x1
x2
x3
x4
x1
x2 y1
x3 y2
x4 y3
x1 y4
H
V=
Surface Area = 4 π R2
Polygon Area
Trapezoids
When A || B then <P = <Q
(B2 + B b + b2) H
V=
3
R
r
H
V=
πH 2
(r + Rr + R2)
3
Solids with common
cross-sections (such as a
barrel) all have similar
volumes
V = π R2 H
V = (ABase)H
H
height is perpendicular
to the base
Solids with similar
cross-sections (such as a
cone) all have similar
volumes
V=
R
(ABase)H
3
R
 Find m: m =
Exponents
am an = am+n
Factor
109
106
103
102
101
10–1
10–2
10–3
10–6
10–9
Example
5 GigaBytes
6 Megahertz
2 kilometers
3 hectograms
5 dekaliters
3 decigrams
2 centimeters
5 milligrams
4 micrograms
5 nanoseconds
Mitered Triangles
a
= am–n
an
a0 = 1
rise y2 – y1
=
run x2 – x1
if the y-intercept is between integers or
off the grid b should be computed using
b = y0 – m x0 where (x0, y0) is any
known point
 Once m & b are found: y = m x + b
Solving Linear Equations
1)
2)
3)
4)
5)
6)
7)
8)
Remove Fractions: mult through by LCD
cancel fractions
Remove ( ): distributive rule; beware of negatives!
Combine Like Terms
Shift variable terms to one side (add/sub)
Shift all other terms to other side
Write as (coefficient) × (variable)
Divide out the Coefficient
Check the Answer
Quadratic Formula
ax2 + bx + c = 0;
m
x=
-b±
b2 − 4ac
2a
Distributive Rule
1
a–n = n
a
a(b ± c) = ab ± ac
(a b)n = (an) (bn)
(b ± c)a = ba ± ca
b±c
b
c
=
±
a
a
a
a n an
( b ) = bn
a = a1/2 = a0.5
ab =
a
=
b
a
π R2 H
3
 Find b: b = y-intercept (i.e. x = 0)
To Graph ax + by = c
Plot the intercepts
 set x = 0, solve for y and plot (0, y)
 set y = 0, solve for x and plot (x, 0
 Draw graph
Symbol
G
M
k
h
da
d
c
m
µ
n
V=
To Find a Line's Equation
To Graph y= mx + b
 Plot (0, b) the y-intercept
 Plot a second point using m by
shifting horizontally by run and
shifting vertically by rise.
 Draw graph
Prefix
Giga
Mega
Kilo
Hecto
Deka
Deci
Centi
Milli
Micro
Nano
Cone
H
height is perpendicular
to the base
Lines
Surface Area =
2πRH + 2πR2
b
a
b
Factoring
ax + bx = (a + b)x
ax + bx − cx = (a + b − c)x
Summary of Trig Relationships
y
Non-trig Relationships using Slope, m
y
m=
x2 + y2 = r2
x
x=
r
1 + m2
y=
rm
1 + m2
(x, y)rect
(r, )polar
y
θ°
a=
2π r (θ in degrees)
360°
a = r θ (θ in radians)

r
Arc Length and Velocity
2πrad = 360°

v = r ω (θ in radians) e.g. (ft/sec = ft × rad/sec)
a
y
x
x
x
Standard Diagram
Trig Relationships
y opp
sin θ = =
r hyp
x adj
cos θ = =
r hyp
y opp
tan θ = =
x adj
Given a right triangle
when θ is known use
these trig relationships
to find a missing side
x = r cos θ
y = r sin θ
θ + α = 90° = π/2rad
tan θ = m
θ = tan-1 m
sin θ = cos α
Given a right triangle
when θ is not known
use these trig
relationships to find
an angle
45° = π/4
180° = π
opp
x
adj
y
opp
( r ) = sin–1 ( hyp )
θ = cos–1
( r ) = cos–1 ( hyp )
θ = tan–1
( x ) = tan–1 ( adj )
60° = π/6
sin2 θ + cos2 θ = 1
cos θ = sin α
y
θ = sin–1
30° = π/12
sin θ
= tan θ
cos θ
Some Examples that do NOT require Trigonometry
Not every triangle requires
trigonometry.
Don't forget to use similar
triangles where possible.
Don’t forget there are 180°
in every triangle.
Two Angles Match
Imbedded Triangles, P||Q
Reflected Triangles, P||Q
Use Similar Triangles
Use Similar Triangles
Use 180° in triangles
Use Similar Triangles
Imbedded Right Triangles
Both legs known,
slope missing
Mitered Right Triangles
Only right triangles can use
the Pythagorean Thm.
Slope only makes sense with
a vertical vs. horizontal
relationship.
Use Similar Triangles
Use Pythagorean Thm.
Use m =
rise
run
Use Similar Triangles
Use Pythagorean Thm.
A right triangle with 2 sides
known, one side missing
Use Pythagorean Thm.
Slope and diagonal known, legs missing
Slope and one leg known, other leg missing
rise
Use m =
run
Use
x=
z
; y=
1 + m2
zm
1 + m2
Examples that do require Trigonometry
The tangent is used when
There is a right triangle
The parts of interest are:
Both legs known, angles missing
one angle and one leg known, other leg missing
Use tan θ = opp/adj
Use θ = tan-1 (opp/adj)
an angle,
both legs,
the slope
Slope known, Standard angle missing
Standard angle known, slope missing
Use tan θ = m
-1
Use θ = tan m
The sine is used when
There is a right triangle
The parts of interest are:
an angle,
its opposite leg,
the diagonal
Angle, opposite leg known,
hypotenuse missing
Angle, hypotenuse known,
leg opposite angle missing
Hypotenuse, leg known;
angle opposite leg missing
Use sin θ = opp/hyp
Use sin θ = opp/hyp
Use θ = sin-1 (opp/hyp)
Angle, adjacent leg known,
hypotenuse missing
Angle, hypotenuse known,
adjacent leg missing
Known sides bracketing
a missing angle
Use cos θ = adj/hyp
Use cos θ = adj/hyp
Use θ = cos-1 (adj/hyp)
The cosine is used when
There is a right triangle
The parts of interest are:
an angle,
its adjacent leg,
the diagonal
When extra information is
known there are multiple
possibilities.
A right triangle with all sides known but angles missing
Use any inverse trig function
Non-Right Triangles use Law of Sines or Law of Cosines
Use Law of Sines with
2 angles + 2 sides with one part missing.
Often must first use sum of angles is 180°.
c
A
B
Law of Sines
a
b
Use Law of Cosines with
3 sides known and missing angles or side-angle-side combo
known with side opposite angle missing
Law of Cosines
sin a sin b sin c
=
=
A
B
C
C
c
A
B
a
b
A2 + B2 – 2ABcos θ = C2
C
cos θ =
A2 + B2 − C2
2AB
Examples
3 Sides known, angles Missing
2 Side-Opposite Angle Combos
with 3 parts known
Use Law of Cosines to find first angle
Many choices after that
side-angle-side combo known
Use Law of Sines to find missing part
Many choices after that
Use Law of Cosines to find third side
angle-angle-side combo known
angle-side-angle combo known
rd
Use 180° in triangle for 3rd angle
Use Law of Sines to find sides
Use 180° in triangle for 3 angle
Use Law of Sines to find sides
Be careful when using Law of Sines
with Obtuse Triangles!
Summary of Basic Trigonometry:
Using Trigonometry in a geometry problem involves triangles. The triangle(s) may be
obvious but sometimes they are imbedded deceptively within a more complicated
geometry. Generally, the task at hand is to use partial information given about the various
triangles to determine all three angles (A) and all three sides (S) of each triangle.
A
S
S
A
A
S
Given only SSS, SAS, ASA, AAS we always have a unique triangle. Thus, the remaining three parts of the
triangle are predetermined. In each of the examples below the remaining three parts (?) can be found using
trigonometry formulas.
?
S
A
?
S
?
S
?
S
SAS
?
S
SSS
?
?
?
?
A
S
ASA
A
A
?
A
?
S
?
AAS
However, if given SSA two possible triangles can occur but only two. Trigonometry can still give us the
unknown parts exactly if we know which of the two cases we have, an acute triangle or an obtuse triangle.
S
S
A
S
S
A
2 possibilities
acute triangle
obtuse triangle
If we have just the three angles known there are an infinite number of similar triangles.
Thus, we must always have at least one side given to exactly determine all parts.
A
?
?
A
With the exception of AAA (similar triangles) knowing three of the six triangle parts allows
us to find the remaining three parts (assuming we have a sketch for the SSA case).
?
A
In the case of right triangles, one angle (the right angle) is already known so we only need two additional parts to
exactly determine all six parts. Right triangles are usually preferable because the relationships are simpler to
work with.
Not every triangle requires trigonometry to determine its components. Often a triangle is oriented so that "slope"
makes sense for one "angle" of the triangle. If a slope is given (or desired) we can often avoid trigonometry and
still determine all the remaining parts of the triangle.
In Trigonometry Applications:
1)
Find triangles inherent and useful to the problem. Label them.
2)
Decide if a triangle is a right-triangle or non-right-triangle.
3)
Determine which relation is required.
4)
Use the relationship to set up an equation.
5)
Solve the equation
6)
Repeat as necessary
Some hints:
Only right triangles can use the Pythagorean Theorem.
Recall there are 180° in a triangle and complementary angles sum to 90°.
Use Sin, Cos or Tan with right triangles when the angle is known.
Use Sin–1, Cos–1 or Tan–1 with right triangles when the angle is the unknown.
When two sides form a "t" use Tan or Tan–1.
Right-triangles which involve the hypotenuse and an angle use either Sin or Cos.
Right-triangles which involve two sides bracketing an angle use Cos.
Right-triangles which involve the hypotenuse and a side opposite an angle use Sin.
Non-right triangles use Law of Sines or Law of Cosines.
When a right triangle is in its standard position the
various relationships are easier to recognize and apply.
standard position with slope standard position with angle
a
c
cm
m= ;
a2 + b2 = c2;
b=
a=
2;
b
1+m
1 + m2
tan θ =
rise
;
run
sin θ =
rise
;
hyp
a = c sin θ
cos θ =
run
hyp
b = c cos θ
When a right triangle is in a random position be careful
when using the trigonometric relationships.
be careful using slope here
a
m = tan θ =
b
opp
a
= tan–1
adj
b
sin θ =
opp b
=
hyp c
opp
adj
; θ = cos–1
hyp
hyp
cos θ =
adj a
=
hyp c
θ = tan–1
θ = sin–1
slope is inapplicable here
opp b
tan θ =
=
adj a
Procedure Summary:
The triangle is a Right Triangle
The triangle is NOT a Right Triangle
If two sides are known, find the third side by
using the Pythagorean Theorem.
If three parts of three sides and one angle are known
use Law of Cosines to find the fourth part.
If one complementary angle is known find
the other using complements sum to 90°.
A2 + B2 – 2AB cos () = C2
If two sides are known find an angle with:
opp
adj
cos () =
A2 + B2 – C2
2AB
opp
 = sin–1 hyp ,  = cos–1 hyp ,  = tan–1 adj .
If one angle and one side are known
find another side using:
opp
adj
opp
sin() = hyp , cos() = hyp , tan () = adj
If three parts of two sides and two angles are known
use Law of Sines to find the fourth part.
sin a
sin b
sin c
A = B = C
Examples
Imbedded Triangles
with P||Q
Reflected Triangles
with P||Q
Use Similar Triangles
Use Similar Triangles
Two Angles Match.
Use 180° in triangles
Use Similar Triangles
Mitered Right Triangles
Use Similar Triangles
Use Pythagorean Thm.
Reference Triangle
3 Sides, Angles Missing
Use Law of Cosines to find
angles
2 Side-Angle Combos
Use Law of Sines to find a
Many choices after that
Side-Angle-Side
Use Law of Cosines to find
third side
Angle- Angle-Side
Use 180° in triangle
Use Law of Sines to find
sides
Angle-Side-Angle
Use 180° in triangle
Use Law of Sines to find
sides
Be careful when using Law of Sines with Obtuse Triangles!