How to find the sine of an angle - ACT Math
Card 0 of 63
Solve for
over the interval 
Solve for over the interval
Substitute x = sinQ and solve the new equation x2 + 3x = –2 by factoring. Be sure to change variables back to Q. As a result, sinQ = –1 or sinQ = –2. This function is bounded between –1 and 1 so sinQ can never be –2 and sinQ is –1 only at 3π/2 or 270 °.
Substitute x = sinQ and solve the new equation x2 + 3x = –2 by factoring. Be sure to change variables back to Q. As a result, sinQ = –1 or sinQ = –2. This function is bounded between –1 and 1 so sinQ can never be –2 and sinQ is –1 only at 3π/2 or 270 °.
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See right triangle ABC. If the length AB is 8 and the length of BC is 6, what is the sine of angle A?
See right triangle ABC. If the length AB is 8 and the length of BC is 6, what is the sine of angle A?
Sine A = Opposite / Hypotenuse = BC / AC
To find AC, use Pythagorean Theorum
AB2 + BC2 = AC2
82 + 62 = AC2
64 + 36 = AC2
100 = AC2
AC = 10
Sine A = BC / AC = 6 / 10 = 0.6
Sine A = Opposite / Hypotenuse = BC / AC
To find AC, use Pythagorean Theorum
AB2 + BC2 = AC2
82 + 62 = AC2
64 + 36 = AC2
100 = AC2
AC = 10
Sine A = BC / AC = 6 / 10 = 0.6
Compare your answer with the correct one above

Triangle
shown is a right triangle. If line
and line
, what is the sine of the angle at
?
Triangle shown is a right triangle. If line
and line
, what is the sine of the angle at
?

Now solve for
using Pythagorean Theorem:






Now solve for using Pythagorean Theorem:
Compare your answer with the correct one above

What is the sine of
?
What is the sine of ?
Sine can be found using the SOH CAH TOA method. For sine we do
.
Sine can be found using the SOH CAH TOA method. For sine we do .
Compare your answer with the correct one above
If
, and if
is an angle between
and
degrees, which of the following equals
?
If , and if
is an angle between
and
degrees, which of the following equals
?
An angle between
and
degrees means that the angle is located in the second quadrant.
The tangent function is derived from taking the side opposite to the angle and dividing by the side adjacent to the angle (
, as shown in the image).

Hence, the side
is
units long and side
is
units high. Therefore, according to Pythagorean Theorem rules, the side
must be
units long (since
).
The sine function is positive in the second quadrant. It is also equivalent to the side opposite the angle (
) divided by the hypotenuse (
).
This makes
.
An angle between and
degrees means that the angle is located in the second quadrant.
The tangent function is derived from taking the side opposite to the angle and dividing by the side adjacent to the angle (, as shown in the image).
Hence, the side is
units long and side
is
units high. Therefore, according to Pythagorean Theorem rules, the side
must be
units long (since
).
The sine function is positive in the second quadrant. It is also equivalent to the side opposite the angle () divided by the hypotenuse (
).
This makes .
Compare your answer with the correct one above
A sine function has a period of
, a
-intercept of
, an amplitude of
and no phase shift. These describe which of these equations?
A sine function has a period of , a
-intercept of
, an amplitude of
and no phase shift. These describe which of these equations?
Looking at this form of a sine function:

We can draw the following conclusions:
because the amplitude is specified as
.
because of the specified period of
since
.
because the problem specifies there is no phase shift.
because the
-intercept of a sine function with no phase shift is
.
Bearing these in mind,
is the only function that fits all four of those.
Looking at this form of a sine function:
We can draw the following conclusions:
because the amplitude is specified as
.
because of the specified period of
since
.
because the problem specifies there is no phase shift.
because the
-intercept of a sine function with no phase shift is
.
Bearing these in mind, is the only function that fits all four of those.
Compare your answer with the correct one above
A sine function has a period of
, a
-intercept of
, an amplitude of
and no phase shift. These describe which of these equations?
A sine function has a period of , a
-intercept of
, an amplitude of
and no phase shift. These describe which of these equations?
Looking at this form of a sine function:

We can draw the following conclusions:
because the amplitude is specified as
.
because of the specified period of
since
.
because the problem specifies there is no phase shift.
because the
-intercept of a sine function with no phase shift is
.
Bearing these in mind,
is the only function that fits all four of those.
Looking at this form of a sine function:
We can draw the following conclusions:
because the amplitude is specified as
.
because of the specified period of
since
.
because the problem specifies there is no phase shift.
because the
-intercept of a sine function with no phase shift is
.
Bearing these in mind, is the only function that fits all four of those.
Compare your answer with the correct one above
Solve for
over the interval 
Solve for over the interval
Substitute x = sinQ and solve the new equation x2 + 3x = –2 by factoring. Be sure to change variables back to Q. As a result, sinQ = –1 or sinQ = –2. This function is bounded between –1 and 1 so sinQ can never be –2 and sinQ is –1 only at 3π/2 or 270 °.
Substitute x = sinQ and solve the new equation x2 + 3x = –2 by factoring. Be sure to change variables back to Q. As a result, sinQ = –1 or sinQ = –2. This function is bounded between –1 and 1 so sinQ can never be –2 and sinQ is –1 only at 3π/2 or 270 °.
Compare your answer with the correct one above

See right triangle ABC. If the length AB is 8 and the length of BC is 6, what is the sine of angle A?
See right triangle ABC. If the length AB is 8 and the length of BC is 6, what is the sine of angle A?
Sine A = Opposite / Hypotenuse = BC / AC
To find AC, use Pythagorean Theorum
AB2 + BC2 = AC2
82 + 62 = AC2
64 + 36 = AC2
100 = AC2
AC = 10
Sine A = BC / AC = 6 / 10 = 0.6
Sine A = Opposite / Hypotenuse = BC / AC
To find AC, use Pythagorean Theorum
AB2 + BC2 = AC2
82 + 62 = AC2
64 + 36 = AC2
100 = AC2
AC = 10
Sine A = BC / AC = 6 / 10 = 0.6
Compare your answer with the correct one above

Triangle
shown is a right triangle. If line
and line
, what is the sine of the angle at
?
Triangle shown is a right triangle. If line
and line
, what is the sine of the angle at
?

Now solve for
using Pythagorean Theorem:






Now solve for using Pythagorean Theorem:
Compare your answer with the correct one above

What is the sine of
?
What is the sine of ?
Sine can be found using the SOH CAH TOA method. For sine we do
.
Sine can be found using the SOH CAH TOA method. For sine we do .
Compare your answer with the correct one above
If
, and if
is an angle between
and
degrees, which of the following equals
?
If , and if
is an angle between
and
degrees, which of the following equals
?
An angle between
and
degrees means that the angle is located in the second quadrant.
The tangent function is derived from taking the side opposite to the angle and dividing by the side adjacent to the angle (
, as shown in the image).

Hence, the side
is
units long and side
is
units high. Therefore, according to Pythagorean Theorem rules, the side
must be
units long (since
).
The sine function is positive in the second quadrant. It is also equivalent to the side opposite the angle (
) divided by the hypotenuse (
).
This makes
.
An angle between and
degrees means that the angle is located in the second quadrant.
The tangent function is derived from taking the side opposite to the angle and dividing by the side adjacent to the angle (, as shown in the image).
Hence, the side is
units long and side
is
units high. Therefore, according to Pythagorean Theorem rules, the side
must be
units long (since
).
The sine function is positive in the second quadrant. It is also equivalent to the side opposite the angle () divided by the hypotenuse (
).
This makes .
Compare your answer with the correct one above
Solve for
over the interval 
Solve for over the interval
Substitute x = sinQ and solve the new equation x2 + 3x = –2 by factoring. Be sure to change variables back to Q. As a result, sinQ = –1 or sinQ = –2. This function is bounded between –1 and 1 so sinQ can never be –2 and sinQ is –1 only at 3π/2 or 270 °.
Substitute x = sinQ and solve the new equation x2 + 3x = –2 by factoring. Be sure to change variables back to Q. As a result, sinQ = –1 or sinQ = –2. This function is bounded between –1 and 1 so sinQ can never be –2 and sinQ is –1 only at 3π/2 or 270 °.
Compare your answer with the correct one above

See right triangle ABC. If the length AB is 8 and the length of BC is 6, what is the sine of angle A?
See right triangle ABC. If the length AB is 8 and the length of BC is 6, what is the sine of angle A?
Sine A = Opposite / Hypotenuse = BC / AC
To find AC, use Pythagorean Theorum
AB2 + BC2 = AC2
82 + 62 = AC2
64 + 36 = AC2
100 = AC2
AC = 10
Sine A = BC / AC = 6 / 10 = 0.6
Sine A = Opposite / Hypotenuse = BC / AC
To find AC, use Pythagorean Theorum
AB2 + BC2 = AC2
82 + 62 = AC2
64 + 36 = AC2
100 = AC2
AC = 10
Sine A = BC / AC = 6 / 10 = 0.6
Compare your answer with the correct one above

Triangle
shown is a right triangle. If line
and line
, what is the sine of the angle at
?
Triangle shown is a right triangle. If line
and line
, what is the sine of the angle at
?

Now solve for
using Pythagorean Theorem:






Now solve for using Pythagorean Theorem:
Compare your answer with the correct one above

What is the sine of
?
What is the sine of ?
Sine can be found using the SOH CAH TOA method. For sine we do
.
Sine can be found using the SOH CAH TOA method. For sine we do .
Compare your answer with the correct one above
If
, and if
is an angle between
and
degrees, which of the following equals
?
If , and if
is an angle between
and
degrees, which of the following equals
?
An angle between
and
degrees means that the angle is located in the second quadrant.
The tangent function is derived from taking the side opposite to the angle and dividing by the side adjacent to the angle (
, as shown in the image).

Hence, the side
is
units long and side
is
units high. Therefore, according to Pythagorean Theorem rules, the side
must be
units long (since
).
The sine function is positive in the second quadrant. It is also equivalent to the side opposite the angle (
) divided by the hypotenuse (
).
This makes
.
An angle between and
degrees means that the angle is located in the second quadrant.
The tangent function is derived from taking the side opposite to the angle and dividing by the side adjacent to the angle (, as shown in the image).
Hence, the side is
units long and side
is
units high. Therefore, according to Pythagorean Theorem rules, the side
must be
units long (since
).
The sine function is positive in the second quadrant. It is also equivalent to the side opposite the angle () divided by the hypotenuse (
).
This makes .
Compare your answer with the correct one above