Trigonometry - ACT Math
Card 0 of 1440
Simplify sec4_Θ_ – tan4_Θ_.
Simplify sec4_Θ_ – tan4_Θ_.
Factor using the difference of two squares: _a_2 – _b_2 = (a + b)(a – b)
The identity 1 + tan2_Θ_ = sec2_Θ_ which can be rewritten as 1 = sec2_Θ_ – tan2_Θ_
So sec4_Θ_ – tan4_Θ_ = (sec2_Θ_ + tan2_Θ_)(sec2_Θ_ – tan2_Θ_) = (sec2_Θ_ + tan2_Θ_)(1) = sec2_Θ_ + tan2_Θ_
Factor using the difference of two squares: _a_2 – _b_2 = (a + b)(a – b)
The identity 1 + tan2_Θ_ = sec2_Θ_ which can be rewritten as 1 = sec2_Θ_ – tan2_Θ_
So sec4_Θ_ – tan4_Θ_ = (sec2_Θ_ + tan2_Θ_)(sec2_Θ_ – tan2_Θ_) = (sec2_Θ_ + tan2_Θ_)(1) = sec2_Θ_ + tan2_Θ_
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If 
, what is 
? Round to the nearest hundredth.
If
Recall that the sine wave is symmetrical with respect to the origin. Therefore, for any value 
, the value for 
is 
. Therefore, if 
is 
, then for 
, it will be 
.
Recall that the sine wave is symmetrical with respect to the origin. Therefore, for any value
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In a right triangle, cos(A) = 
. What is sin(A)?
In a right triangle, cos(A) =
In a right triangle, for sides a and b, with c being the hypotenuse, 
. Thus if cos(A) is 
, then c = 14, and the side adjacent to A is 11. Therefore, the side opposite of angle A is the square root of 
, which is 
Since sin is 
, sin(A) is 
.
In a right triangle, for sides a and b, with c being the hypotenuse,
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A function with period P will repeat on intervals of length P, and these intervals are referred to as periods.
Find the period of
.
A function with period P will repeat on intervals of length P, and these intervals are referred to as periods.
Find the period of
For the function
the period is equal to
in this case
which reduces to 
.
For the function
the period is equal to
in this case
which reduces to
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If angle A measures 30 degrees and the hypotenuse is 4, what is the length of AB in the given right triangle?
If angle A measures 30 degrees and the hypotenuse is 4, what is the length of AB in the given right triangle?
Cosine A = Adjacent / Hypotenuse = AB / AC = AB / 4
Cosine A = AB / 4
Cos (30º) = √3 / 2 = AB / 4
Solve for AB
√3 / 2 = AB / 4
AB = 4 * (√3 / 2) = 2√3
Cosine A = Adjacent / Hypotenuse = AB / AC = AB / 4
Cosine A = AB / 4
Cos (30º) = √3 / 2 = AB / 4
Solve for AB
√3 / 2 = AB / 4
AB = 4 * (√3 / 2) = 2√3
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What is the value of 
in the right triangle above? Round to the nearest hundredth.
What is the value of
Recall that the tangent of an angle is the ratio of the opposite side to the adjacent side of that triangle. Thus, for this triangle, we can say:
Solving for 
, we get:
or 
Recall that the tangent of an angle is the ratio of the opposite side to the adjacent side of that triangle. Thus, for this triangle, we can say:
Solving for
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In the right triangle shown above, let 
, 
, and 
. What is the value of 
Reduce all fractions.
In the right triangle shown above, let
First we need to find the value of 
. Use the mnemonic SOH-CAH-TOA which stands for:
.
Now we see at point 
we are looking for the opposite and adjacent sides, which are 
and 
respectively.
Thus we get that
and plugging in our values and reducing yields:
First we need to find the value of
Now we see at point
Thus we get that
and plugging in our values and reducing yields:
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First, solve for side MN. Tan(30°) = MN/16√3, so MN = tan(30°)(16√3) = 16. Triangle LMN and MNO are similar as they're both 30-60-90 triangles, so we can set up the proportion LM/MN = MN/NO or 16√3/16 = 16/x. Solving for x, we get 9.24, so the closest whole number is 9.
First, solve for side MN. Tan(30°) = MN/16√3, so MN = tan(30°)(16√3) = 16. Triangle LMN and MNO are similar as they're both 30-60-90 triangles, so we can set up the proportion LM/MN = MN/NO or 16√3/16 = 16/x. Solving for x, we get 9.24, so the closest whole number is 9.
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Josh is at the state fair when he decides to take a helicopter ride. He looks down at about a 35 ° angle of depression and sees his house. If the helicopter was about 250 ft above the ground, how far does the helicopter have to travel to be directly above his house?
sin 35 ° = 0.57 cos 35 ° = 0.82 tan 35 ° = 0.70
Josh is at the state fair when he decides to take a helicopter ride. He looks down at about a 35 ° angle of depression and sees his house. If the helicopter was about 250 ft above the ground, how far does the helicopter have to travel to be directly above his house?
sin 35 ° = 0.57 cos 35 ° = 0.82 tan 35 ° = 0.70
The angle of depression is the angle formed by a horizontal line and the line of sight looking down from the horizontal.
This is a right triangle trig problem. The vertical distance is 250 ft and the horizontal distance is unknown. The angle of depression is 35°. We have an angle and two legs, so we use tan Θ = opposite ÷ adjacent. This gives an equation of tan 35° = 250/d where d is the unknown distance to be directly over the house.
The angle of depression is the angle formed by a horizontal line and the line of sight looking down from the horizontal.
This is a right triangle trig problem. The vertical distance is 250 ft and the horizontal distance is unknown. The angle of depression is 35°. We have an angle and two legs, so we use tan Θ = opposite ÷ adjacent. This gives an equation of tan 35° = 250/d where d is the unknown distance to be directly over the house.
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Consider the triangle 
where 
. Find 
to the nearest decimal place.
Note: The triangle is not necessarily to scale
Consider the triangle where
Note: The triangle is not necessarily to scale
To solve this equation, it is best to remember the mnemonic SOHCAHTOA which translates to Sin = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent. Looking at the problem statement, we are given an angle and the side opposite of the angle, and we are looking for the side adjacent to the angle. Therefore, we will be using the TOA part of the mnemonic. Inserting the values given in the problem statement, we can write 
. Rearranging, we get 
. Therefore 
To solve this equation, it is best to remember the mnemonic SOHCAHTOA which translates to Sin = Opposite / Hypotenuse, Cosine = Adjacent / Hypotenuse, and Tangent = Opposite / Adjacent. Looking at the problem statement, we are given an angle and the side opposite of the angle, and we are looking for the side adjacent to the angle. Therefore, we will be using the TOA part of the mnemonic. Inserting the values given in the problem statement, we can write
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A piece of wire is tethered to a 
building at a 
angle. How far back is this wire from the bottom of said building? Round to the nearest inch.
A piece of wire is tethered to a
Begin by drawing out this scenario using a little right triangle:
Note importantly: We are looking for 
as the the distance to the bottomof the building. Now, this is not very hard at all! We know that the tangent of an angle is equal to the ratio of the side adjacent to that angle to the_opposite_ side of the triangle. Thus, for our triangle, we know:
Using your calculator, solve for 
:
This is 
. Now, take the decimal portion in order to find the number of inches involved.
Thus, rounded, your answer is 
feet and 
inches.
Begin by drawing out this scenario using a little right triangle:
Note importantly: We are looking for
Using your calculator, solve for
This is
Thus, rounded, your answer is
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In a given right triangle 
, leg 
and 
. Using the definition of 
, find the length of leg 
. Round all calculations to the nearest tenth.
In a given right triangle
In right triangles, SOHCAHTOA tells us that 
, and we know that 
and leg
. Therefore, a simple substitution and some algebra gives us our answer.
Use a calculator or reference to approximate cosine.
Isolate the variable term.
Thus, 
.
In right triangles, SOHCAHTOA tells us that
Thus,
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In a given right triangle , leg 
and 
. Using the definition of 
, find the length of leg . Round all calculations to the nearest hundredth.
In a given right triangle
In right triangles, SOHCAHTOA tells us that 
, and we know that 
and hypotenuse 
. Therefore, a simple substitution and some algebra gives us our answer.
Use a calculator or reference to approximate cosine.
Isolate the variable term.
Thus, 
.
In right triangles, SOHCAHTOA tells us that
Thus,
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In a given right triangle , leg 
and 
. Using the definition of 
, find the length of leg . Round all calculations to the nearest tenth.
In a given right triangle
In right triangles, SOHCAHTOA tells us that 
, and we know that 
and hypotenuse 
. Therefore, a simple substitution and some algebra gives us our answer.
Use a calculator or reference to approximate cosine.
Isolate the variable term.
Thus, 
.
In right triangles, SOHCAHTOA tells us that
Thus,
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In the above triangle, 
and 
. Find 
.
In the above triangle,
With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the opposite and adjacent sides of the triangle with relation to the angle. With this information, we can use the tangent function to find the angle.
With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the opposite and adjacent sides of the triangle with relation to the angle. With this information, we can use the tangent function to find the angle.
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For the above triangle, 
and 
. Find 
.
For the above triangle,
With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the opposite and adjacent sides of the triangle with relation to the angle. With this information, we can use the tangent function to find the angle.
With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the opposite and adjacent sides of the triangle with relation to the angle. With this information, we can use the tangent function to find the angle.
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For the above triangle, 
and 
. Find 
.
For the above triangle,
With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the opposite and adjacent sides of the triangle with relation to the angle. With this information, we can use the tangent function to find the angle.
With right triangles, we can use SOH CAH TOA to solve for unknown side lengths and angles. For this problem, we are given the opposite and adjacent sides of the triangle with relation to the angle. With this information, we can use the tangent function to find the angle.
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A laser is placed at a distance of 
from the base of a building that is 
tall. What is the angle of the laser (presuming that it is at ground level) in order that it point at the top of the building?
A laser is placed at a distance of
You can draw your scenario using the following right triangle:
Recall that the tangent of an angle is equal to the ratio of the opposite side to the adjacent side of the triangle. You can solve for the angle by using an inverse tangent function:
or 
.
You can draw your scenario using the following right triangle:
Recall that the tangent of an angle is equal to the ratio of the opposite side to the adjacent side of the triangle. You can solve for the angle by using an inverse tangent function:
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What is the value of 
in the right triangle above? Round to the nearest hundredth of a degree.
What is the value of
Recall that the tangent of an angle is equal to the ratio of the opposite side to the adjacent side of the triangle. You can solve for the angle by using an inverse tangent function:
or 
.
Recall that the tangent of an angle is equal to the ratio of the opposite side to the adjacent side of the triangle. You can solve for the angle by using an inverse tangent function:
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What is the tangent of the angle formed between the origin and the point 
if that angle is formed with one side of the angle beginning on the 
-axis and then rotating counter-clockwise to ? Round to the nearest hundredth.
What is the tangent of the angle formed between the origin and the point
Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the 
-axis as your reference point for your angle. (Hence, it is called the "reference angle.")
Now, it is easiest to think of this like you are drawing a little triangle in the fourth quadrant of the Cartesian plane. It would look like:
So, the tangent of an angle is:
or, for your data, 
or 
. However, since 
is in the fourth quadrant, your value must be negative. (The tangent function is negative in that quadrant.) This makes the correct answer 
.
Recall that when you calculate a trigonometric function for an obtuse angle like this, you always use the
Now, it is easiest to think of this like you are drawing a little triangle in the fourth quadrant of the Cartesian plane. It would look like:
So, the tangent of an angle is:
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