Sat math Essentials

e.
22
3
Answer
c.
In a 30-60-90 triangle, the leg opposite the 30° angle 
half the length of the hypotenuse. The
hypotenuse is 22, so the leg opposite the 30° angle 
11. The leg opposite the 60° angle 
3
the
length of the other leg. The other leg 
11, so the leg opposite the 60° angle 
11
3
.
60
°
22
30
°
x
y
60
°
12
30
°
y
x
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Triangle Trigonometry 
There are special ratios we can use when working with right triangles. They are based on the trigonometric func-
tions called 
sine
,
cosine
, and 
tangent
.
For an angle,
, within a right triangle, we can use these formulas:
sin 
hy
o
p
p
o
p
t
o
e
s
n
i
u
te
se
cos 
hy
a
p
d
o
ja
t
c
e
e
n
n
u
t
se
tan 
o
ad
p
j
p
a
o
c
s
e
i
n
te
t
The popular mnemonic to use to remember these formulas is 
SOH CAH TOA
.
SOH
stands for 
S
in:
O
pposite/
H
ypotenuse
CAH
stands for 
C
os:
A
djacent/
H
ypotenuse
TOA
stands for 
T
an:
O
pposite/
A
djacent
Although trigonometry is tested on the SAT, all SAT trigonometry questions can also be solved using geom-
etry (such as rules of 45-45-90 and 30-60-90 triangles), so knowledge of trigonometry is not essential. But if you
don’t bother learning trigonometry, be sure you understand triangle geometry completely.
oppo
s
ite
hypotenu
s
e
adjacent
hypotenu
s
e
oppo
s
ite
adjacent
To find 
s
in 
... 
To find co
s
... 
To find tan 
... 
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TRIG VALUES OF SOME COMMON ANGLES
SIN
COS
TAN
30°
1
2
45°
1
60°
1
2
3
3
2
2
2
2
2
3
3
3
2

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Example
First, let’s solve using trigonometry:
We know that cos 45°
, so we can write an equation:
hy
a
p
d
o
ja
t
c
e
e
n
n
u
t
se
1
x
0
Find cross products.
2 
10 
x
2
Simplify.
20 
x
2
x
Now, multiply 
by 
(which equals 1), to remove the 
2
from the denominator.
x
x
10
2
x
Now let’s solve using rules of 45-45-90 triangles, which is a lot simpler:
The length of the hypotenuse 
2
the length of a leg of the triangle. Therefore, because the leg is 10, the
hypotenuse is 
2
10 
10
2
.
20
2
2
20
2
2
2
2
2
20
2
20
2
2
2
2
2
2
2
45°
x
10
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C i r c l e s
A 
circle
is a closed figure in which each point of the circle is the same distance from the center of the circle.
Angles and Arcs of a Circle
■
An
arc
is a curved section of a circle.
■
A 
minor arc
is an arc less than or equal to 180°. A 
major arc
is an arc greater than or equal to 180°.
■
A 
central angle
of a circle is an angle with its vertex at the center and sides that are radii. Arcs have the same
degree measure as the central angle whose sides meet the circle at the two ends of the arc.
Central 
A
ngle
Major Arc
Minor Arc
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Length of an Arc
To find the length of an arc, multiply the circumference of the circle, 2
π
r
, where 
r
the radius of the circle, by
the fraction 
36
x
0
, with 
x
being the degree measure of the central angle:
2
π
r
36
x
0
2
3
π
6
r
0
x
1
π
8
rx
0
Example
Find the length of the arc if
x
90 and 
r
56.
L
1
π
8
rx
0
L
π
(5
1
6
8
)
0
(90)
L
π
(
2
56)
L
28
π
The length of the arc is 28
π
.
Practice Question
If
x
32 and 
r
18, what is the length of the arc shown in the figure above?
a.
16
5
π
b.
32
5
π
c.
36
π
d.
28
5
8
π
e.
576
π
x°
r
r
r
x°
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Answer
a.
To find the length of an arc, use the formula 
1
π
8
rx
0
, where 
r
the radius of the circle and 
x
the meas-
ure of the central angle of the arc. In this case,
r
18 and 
x
32.
1
π
8
rx
0
π
(1
1
8
8
)
0
(32)
π
1
(3
0
2)
π
(
5
16)
16
5
π
Area of a Sector
A 
sector
of a circle is a slice of a circle formed by two radii and an arc.
To find the area of a sector, multiply the area of a circle,
π
r
2
, by the fraction 
36
x
0
, with 
x
being the degree meas-
ure of the central angle:
π
3
r
6
2
0
x
.
Example
Given 
x
120 and 
r
9, find the area of the sector:
A
π
3
r
6
2
0
x
A
π
(9
2
3
)
6
(
0
120)
A
π
(
3
9
2
)
A
81
3
π
A
27
π
The area of the sector is 27
π
.
x
°
r
r
sector
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Practice Question
What is the area of the sector shown above?
a.
4
3
9
6
π
0
b.
7
3
π
c.
49
3
π
d.
280
π
e.
5,880
π
Answer
c.
To find the area of a sector, use the formula 
π
3
r
6
2
0
x
, where 
r
the radius of the circle and 
x
the measure
of the central angle of the arc. In this case,
r
7 and 
x
120.
π
3
r
6
2
0
x
π
(7
2
3
)
6
(
0
120)
π
(49
3
)
6
(
0
120)
π
(
3
49)
49
3
π
Tangents
A 
tangent
is a line that intersects a circle at one point only.
tangent
point of intersection
120
°
7
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There are two rules related to tangents:
1.
A radius whose endpoint is on the tangent is always perpendicular to the tangent line.
2.
Any point outside a circle can extend exactly two tangent lines to the circle. The distances from the origin
of the tangents to the points where the tangents intersect with the circle are equal.
Practice Question
What is the length of
A
B
in the figure above if
B
C
is the radius of the circle and 
A
B
is tangent to the circle?
a.
3
b.
3
2
c.
6
2
d.
6
3
e.
12
A
B
6
3
0°
C
AB = AC
— —
B
C
A
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Answer
d.
This problem requires knowledge of several rules of geometry. A tangent intersects with the radius of a
circle at 90°. Therefore,
Δ
ABC
is a right triangle. Because one angle is 90° and another angle is 30°,
then the third angle must be 60°. The triangle is therefore a 30-60-90 triangle.
In a 30-60-90 triangle, the leg opposite the 60° angle is 
3
the leg opposite the 30° angle. In
this figure, the leg opposite the 30° angle is 6, so 
A
B
, which is the leg opposite the 60° angle, must be
6
3
.
P o l y g o n s
A 
polygon
is a closed figure with three or more sides.
Example
Terms Related to Polygons
■
A 
regular
(or equilateral) polygon has sides that are all equal; an 
equiangular
polygon has angles that are all
equal. The triangle below is a regular and equiangular polygon:
■
Vertices are corner points of a polygon. The vertices in the six-sided polygon below are:
A
,
B
,
C
,
D
,
E
, and 
F
.
B
C
F
A
D
E
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■
A 
diagonal
of a polygon is a line segment between two non-adjacent vertices. The diagonals in the polygon
below are line segments 
A
C
,
A
D
,
A
E
,
B
D
,
B
E
,
B
F
,
C
E
,
C
F
, and 
D
F
.
Quadrilaterals
A 
quadrilateral
is a four-sided polygon. Any quadrilateral can be divided by a diagonal into two triangles, which
means the sum of a quadrilateral’s angles is 180°
180°
360°.
Sums of Interior and Exterior Angles
To find the sum of the 
interior angles
of any polygon, use the following formula:
S
180(
x
2), with 
x
being the number of sides in the polygon.
Example
Find the sum of the angles in the six-sided polygon below:
S
180(
x
2)
S
180(6 
2)
S
180(4)
S
720
The sum of the angles in the polygon is 720°.