In this article we have covered wide variety of Right Triangle Trigonometry Worksheets that are suitable for middle schoolers. For each section, we have given the methodology that can be used to solve the problems in the worksheet.

Feel free to download and print (for personal use) and try these intuitive trigonometry problems.

### Applications of Right Triangle Trigonometry Worksheet

To solve these problems, you will have to first learn the concept of the Pythagorean Theorem and the law of tangents.

** The Pythagorean Theorem –**Is about the relationship between the three sides of a right-angle triangle. So, if ABC is a right-angle triangle in which the three sides are AB, BC, and AC, AB

^{2 }+ BC

^{2}=AC

^{2}. Here, AC is the hypotenuse-the longest side of the right angle triangle.

You are needed to use the above formula to determine the unknown side of the right-angled triangle when two sides are already given.

__Law of tangents __**–**It is the relationship between any two sides and the two angles opposite to these two sides of a right-angled triangle ABC. Remember, the law of tangent applies to only right-angle triangles.

Again tangent of a given angle of a right-angled triangle is the ratio of its opposite side to its adjacent side.

So, here tanC=AB/BC

Example of finding the length of a side.

Calculate the length of x and y of two right-angle triangles in the figure below that share two vertices and one common straight-line base. Here the length of one side AB is 12, D=60^{0} and angle C=30^{0}.

To solve the above problem, you have to do the following steps.

Step 1. For triangle ABD, you already have AB=12, B=90^{0} and D=60^{0}

So, tan 60^{0 }=opposite side/adjacent side

=AB/BD

=12/BD

=12/x

Since it is known from the trigonometry table that tan 60^{0 }=√3, ^{ }you can write

√3=12/x

So, x=12/√3

=3×4/√3

=√3x√3 x4/√3

=4√3

Step 2 For triangle ABC, you already have AB=12, B=90^{0} and C=30^{0}

So, tan 30^{0 }=opposite side/adjacent side

=AB/BC

=12/(BD +DC)

=12/(x+y)

Since it is known from the trigonometry table that tan 30^{0 }=1/√3, ^{ }you can write

1/√3=12/(4√3 +y)

So, (4√3 +y)=12x√3

y =12√3-4√3

=√3(12-4)

=8√3

Now since √3=1.732, you have

x=4×1.732

=6.928=7 (approximately )

and y=8×1.732

=13.85=14 (approximately)

**Worksheet for you to try :**

### Basic Right Triangle Trigonometry Worksheets

To solve these problems, you have to use the law of tangents, sines, and cosines as per the given adjacent side, opposite side and the hypotenuse.

Example of finding the angle of a right angle triangle when the adjacent side and the hypotenuse are already given

Calculate the angle x of the right-angle triangle in the figure below. Here, the length of one side BC =3 and the hypotenuse AC=6.

To solve the above problem, you have to do the following steps.

Since you have the length of the adjacent side of the angle C and the hypotenuse, you have to use the law of cosine.

So, cos x=adjacent side/hypotenuse

=BC/AC

=3/6

=1/2

Now, as we know cos 60^{0} =1/2, so , x=60^{0}

Example of finding the angle of a right-angle triangle when the opposite side and the hypotenuse are already given

Calculate the angle x of the right-angle triangle in the figure below. Here the length of one side BC =14 and the hypotenuse AC=21.

To solve the above problem, you have to do the following steps.

Since, you have the length of the opposite side of the angle A and the hypotenuse, you have to use the law of sine

So, sin x=opposite side/hypotenuse

=BC/AC

=14/21

=2/3

=0.666

Now as sin 41.8^{0} =0.666, so, x=41.8^{0}

**Worksheets to try:**

### Practice Worksheet – Right Triangle Trigonometry

Example of finding sin 45 degrees and cos 45 degrees of the right angle triangle in fraction form when all the three sides are given.

Sin 45^{0 }=opposite side/hypotenuse

= 4/7.2

=40/72

=10/18

=5/9

Cos45^{0 }=adjacent side/hypotenuse

= 6/7.2

=60/72

=10/12

=5/6

### Precalculus Right Triangle Trigonometry Worksheet

Example A swimmer is 210 meter below the surface of the ocean and begins to descend at an angle of 30 degrees from the vertical. How far will be the swimmer travel before he breaks the surface of the water.

You know here Sin 30^{0 }=opposite side/hypotenuse

1/2 =210/hypotenuse [As sin30 degree=1/2]

So hyotenuse =210 ÷0.5

=420

So, the swimmer travels 420 meter before he breaks the ocean surface.

### Right Triangle Trigonometry – Angle of Elevation and Depression Worksheet

To solve the problems in the worksheet, you need to first know the concept of the angle of elevation and the angle of depression.

** The angle of elevation**-The angle formed when an observer looks at an object above his horizontal line of sight. For example, if you stand on a plateau and look at the peak of a nearby mountain, an angle of elevation is formed.

** The angle of depression**-The angle formed when an observer looks down at an object below his horizontal line of sight. For example, if you stand on a plateau and look at a house in the plains, an angle of depression is formed.

Example- A bird is sitting on an iceberg 100 feet above the water. If a sea lion in water is 220 feet from the base of the iceberg, find the angle of depression.

Let x is the angle of elevation that the sea lion at a distance of 220 meter from the base of the iceberg makes when he looks at the bird sitting on the iceberg at a vertical height of 100 meters.

From the above figure, it becomes apparent that tan x^{0 }=adjacent side /opposite side

=220/100

=11/5

tan 66^{0}=2.2

Now as tan 66^{0}=2.2,you have angle of elevation x= 66^{0}

Next, we know that the sum of all three angles of a triangle equals to 180^{0}

So, the remaining angle will be =180^{0 }–(66^{0 }+ 90^{0}) = 180^{0 }– 156^{0}=24^{0}

Hence, the angle of depression = 90^{0 }– 24^{0}=66^{0}

Example A 25 feet ladder leans against a house so that the base of the ladder is 7 feet from the base of the house. What will be the angle of elevation of the ladder.

cos x=adjacent side/hypotenuse

=7/14

=1/2 [cos 60^{0}=1/2]

So, angle of elevation of the ladder= 60^{0}

### Special Right Triangle Trigonometry Worksheet

Example-Half of an equilateral triangle is often called “30-60” right or “30-60-90” triangle. Explain why it is called with that name?

Since, an equilateral triangle can be split into two right angle triangles with the remaining angles being 30 degrees and 60 degrees, half of an equilateral triangle is often called “30-60” right or “30-60-90” triangle. Let us prove this with the help of the law of sine and cosine.

Let ABC is an equilateral triangle with sides x, y and z where x=y=z.

Now you draw a perpendicular from vertex A on the side BC, you will have two right angle triangles-ABD and ACD with angle D=90^{0}.

Next, as per the law of cosine, you know in the right-angle triangle ADC

cos C=adjacent side/hypotenuse

=DC/AC

=Half of the side z/y

=Half of the side y/y [since y=z]

=y/2 ÷ y

=1/2

Now from the trigonometry table, you know that cos 60^{0}=1/2

So, ∠C=60^{0}

Similarly, as per the law of sine, you know that in the right-angle triangle ADC

Sin A=opposite side/hypotenuse

= DC/AC

=Half of the side z/y

=Half of the side y/y [since y=z]

=y/2 ÷ y

=1/2

Now from the trigonometry table, you know that sin30^{0}=1/2

So, ∠A=30^{0}

Hence, you have in the right-angle triangle ACD, A=30^{0}, C=60^{0}, and D=90^{0}

Similarly, in the right-angle triangle ABD, A=30^{0}, B=60^{0}, and D=90^{0}