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Trigonometry Word Problems

Applications of trigonometric ratios.

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Triangles in Applied Problems

A community garden is being built in your neighborhood. The garden will be triangular in shape and a fence will surround the garden. Two sides of the garden will be 33 feet and 24 feet. The angle between those two sides will measure \begin{align*}62^\circ\end{align*}. Find the area and perimeter of the garden.

Triangle Summary

There are many different problems you can solve with your knowledge of triangles and trigonometry. Here is a summary of all the key facts and formulas that will be helpful. 

TRIANGLE SUMMARY:

  • The sum of the measures of the three angles in a triangle is \begin{align*}180^\circ\end{align*}
  • In 30-60-90 right triangles the sides are in the ratio \begin{align*}1:\sqrt{3}:2\end{align*}
  • In 45-45-90 right triangles the sides are in the ratio \begin{align*}1:1: \sqrt{2}\end{align*}
  • The Pythagorean Theorem states that for a right triangle with legs \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and hypotenuse \begin{align*}c\end{align*}, \begin{align*}a^2+b^2=c^2\end{align*}
  • SOH CAH TOA is a mnemonic device to help you remember the three trigonometric ratios:

\begin{align*}\sin \theta=\frac{\text{opposite leg}}{\text{hypotenuse}} \quad \cos \theta=\frac{\text{adjacent leg}}{\text{hypotenuse}} \quad \tan \theta=\frac{\text{opposite leg}}{\text{adjacent leg}}\end{align*}

  • The Law of Sines\begin{align*}\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\end{align*} (watch out for the SSA case)
  • The Law of Cosines: \begin{align*}c^2=a^2+b^2-2ab \cos C\end{align*}
  • The area of a triangle is \begin{align*}\frac{1}{2} bh\end{align*} or \begin{align*}\frac{1}{2} ab \sin C\end{align*}

 

Real-World Application: Boating 

If a boat travels 4 miles SW (southwest) and then 3 miles NNW (north-northwest), how far away is the boat from its starting point?

First, think about what southwest and north-northwest mean. The picture below shows the four basic directions of north, south, east, and west. It also shows southwest and northwest. North-northwest is directly in between northwest and north. Make sure you understand where the angles in the picture came from.

Next, draw a picture of the situation.

Note that the angle between the SW direction and the NNW direction is \begin{align*}22.5^\circ+45^\circ=67.5^\circ\end{align*}.

Now make a plan for how you will solve. Look to see what is given and what you are looking for and think about what method or technique would be helpful. In this situation, you have two sides and an included angle and are looking for the third side. You can use the Law of Cosines.

Now that you have a plan, you can solve the problem. For the Law of Cosines, the 3 and 4 are the values for \begin{align*}a\end{align*} and \begin{align*}b\end{align*} and \begin{align*}67.5^\circ\end{align*} is the value for \begin{align*}\angle C\end{align*}\begin{align*}x\end{align*} is the side across from the angle, so it is \begin{align*}c\end{align*}.

\begin{align*}a^2+b^2-2ab \cos C &= c^2\\ 3^2+4^2-2(3)(4) (\cos 67.5) &= x^2\\ 9+16-9.18 &= x^2\\ 15.82 &= x^2\\ x & \approx 3.98\end{align*}

The boat is 3.98 miles from its starting place.

Real-World Application: Observation Points 

1. From the fourth story of a building (65 feet) Mark observes a car moving towards the building driving on the street below. If the angle of depression of the car changes from \begin{align*}15^\circ\end{align*} to \begin{align*}45^\circ\end{align*} while he watches, how far did the car travel?

The angle of depression is the angle at which you view an object below the horizon. Start by making a detailed picture of the situation and labeling what you know.

Note that \begin{align*}15^\circ+30^\circ=45^\circ\end{align*}, the angle of depression for the ending location of the car.

Now make a plan for how you will solve. Look to see what is given and what you are looking for and think about what method or technique would be helpful. In this situation, you have two right triangles (the smaller brown triangle and the larger blue triangle). In each case, you know a side and an angle. You are looking for a portion of one of the sides of these triangles \begin{align*}(x)\end{align*}.

You can use trigonometric ratios to find the missing sides of these triangles to help you to find the length of \begin{align*}x\end{align*}.

First look at the small brown triangle. \begin{align*}y\end{align*} is opposite the \begin{align*}45^\circ\end{align*} angle and 65 is adjacent to the \begin{align*}45^\circ\end{align*} angle. This is a tangent relationship (or you could use 45-45-90 triangle ratios):

\begin{align*}\tan 45^\circ &= \frac{y}{65}\\ y &= 65 \tan 45^\circ\\ y &= 65 \ ft\end{align*}

Now look at the larger right triangle. \begin{align*}x+y\end{align*} is opposite the \begin{align*}75^\circ\end{align*} angle \begin{align*}(30^\circ+45^\circ=75^\circ)\end{align*} and 65 is adjacent to the \begin{align*}75^\circ\end{align*} angle. Again you can use tangent.

\begin{align*}\tan 75^\circ &= \frac{x+y}{65}\\ 65 \tan 75^\circ &= x+y\\ x+y & \approx 242.58 \ ft\end{align*}

Since \begin{align*}y=65 \ ft\end{align*}, \begin{align*}x\end{align*}, must equal \begin{align*}242.58-65=177.58 \ ft\end{align*}.

The car traveled 177.58 feet.

2. Karen is 5.5 feet tall and looks up at a \begin{align*}40^\circ\end{align*} angle to see the top of the flagpole in front of a building. She is standing 40 feet from the flagpole. How tall is the flagpole?

Again, start by making a detailed picture of the situation and labeling what you know.

Now, make a plan for how you will solve. Look to see what is given and what you are looking for and think about what method or technique would be helpful. In this situation, you have a right triangle. You know an angle and a side within the right triangle (\begin{align*}40^\circ\end{align*} and 40 ft). You are looking for another side of the triangle \begin{align*}(x)\end{align*}.

Think of the flagpole as being made up of two pieces. The first piece is Karen's height of 5.5 feet. The next piece is the rest of the flagpole that is taller than Karen (\begin{align*}x\end{align*} in the picture above). \begin{align*}x\end{align*} is opposite the \begin{align*}40^\circ\end{align*} angle and 40 ft is adjacent to the \begin{align*}40^\circ\end{align*} angle. This is a tangent relationship.

\begin{align*}\tan 40^\circ &= \frac{x}{40}\\ x &= 40 \tan 40^\circ\\ x & \approx 33.56 \ ft\end{align*}

Therefore, the complete height of the flagpole is approximately \begin{align*}33.56+5.5=39.06 \ feet\end{align*}.

Examples

Example 1

Earlier, you were asked to find the area and perimeter of the garden. 

A community garden is being built in your neighborhood. The garden will be triangular in shape and a fence will surround the garden. Two sides of the garden will be 33 feet and 24 feet. The angle between those two sides will measure \begin{align*}62^\circ\end{align*}. Find the area and perimeter of the garden.

Start by drawing a picture and carefully labeling everything that you know.

To find the area of the garden you can use the sine area formula.

\begin{align*}A &= \frac{1}{2} (33)(24) \sin 62^\circ\\ A & \approx 349.6 \ ft^2\end{align*}

In order to find the perimeter of the garden you need to know the length of the third side. In this situation you know two sides and an included angle, so you can use the Law of Cosines to find the length of the third side.

\begin{align*}33^2+24^2-2(33) (24) \cos 62^\circ &= x^2\\ 1089+576-743.64 &= x^2\\ 921.36 &= x^2\\ x & \approx 30.35 \ ft\end{align*}

Now that you know the length of the third side, you can find the perimeter of the garden.

\begin{align*}P=33+24+30.35=87.35 \ ft\end{align*}

A surveyor wants to find the distance from points \begin{align*}A\end{align*} and \begin{align*}B\end{align*} to an inaccessible point \begin{align*}C\end{align*}. These three points form a triangle. Standing at point \begin{align*}A\end{align*}, he finds \begin{align*}m \angle A\end{align*} in the triangle is equal to \begin{align*}60^\circ\end{align*}. Standing at point \begin{align*}B\end{align*}, he finds \begin{align*}m \angle B\end{align*} in the triangle is equal to \begin{align*}55^\circ\end{align*}. He measures the distance from point \begin{align*}A\end{align*} to point \begin{align*}B\end{align*} and finds it to be 350 feet. Find the distance from points \begin{align*}A\end{align*} and \begin{align*}B\end{align*} to point \begin{align*}C\end{align*}

Example 2

Draw a picture of this situation.

Here is a picture:

Example 3

Make a plan: what method(s) or technique(s) can you use to solve this problem?

You can find the measure of angle \begin{align*}C\end{align*} using the fact that the sum of the measures of the three angles in a triangle is \begin{align*}180^\circ\end{align*}. Then, you will know all the angles and one side. You could then use the Law of Sines to set up equations to find \begin{align*}AC\end{align*} and \begin{align*}BC\end{align*}.

Example 4

Solve the problem. 

\begin{align*}m \angle C=65^\circ\end{align*}. Now, use the Law of Sines twice:

To find \begin{align*}AC\end{align*}:

\begin{align*}\frac{\sin 65^\circ}{350} &= \frac{\sin 55^\circ}{AC}\\ AC &= \frac{350 \sin 55^\circ}{\sin 65^\circ}\\ AC &\approx 316.34 \ ft\end{align*}

To find \begin{align*}BC\end{align*}:

\begin{align*}\frac{\sin 65^\circ}{350} &= \frac{\sin 60^\circ}{BC}\\ AC &= \frac{350 \sin 60^\circ}{\sin 65^\circ}\\ AC &\approx 334.44 \ ft\end{align*}

The distance from \begin{align*}A\end{align*} to \begin{align*}C\end{align*} is approximately 316 feet and the distance from \begin{align*}B\end{align*} to \begin{align*}C\end{align*} is approximately 334 feet.

Review

The angle of depression of a boat in the distance from the top of a lighthouse is \begin{align*}25^\circ\end{align*}. The lighthouse is 200 feet tall. Find the distance from the base of the lighthouse to the boat. 

1. Draw a picture of this situation.

2. Make a plan: what method(s) or technique(s) can you use to solve this problem?

3. Solve the problem. 

A pilot is flying due west and gets word that a major storm is in her path. She turns the plane \begin{align*}40^\circ\end{align*} to the left of her intended course and continues the flying. After passing the storm, she turns \begin{align*}50^\circ\end{align*} to the right and flies until she has returned to her original flight path. At this point she is 75 miles from where she left her original path when she first made a turn. How much further did the pilot fly as a result of the detour?

4. Draw a picture of this situation.

5. Make a plan: what method(s) or technique(s) can you use to solve this problem?

6. Solve the problem. 

A new bridge is being built across a river in your town. You want to figure out how long the bridge will be. You find two points on one side of the river that are 30 feet apart. These two points with the point at the end of the bridge on the other side of the river form a triangle. You stand at each of the two points on your side of the river and measure the angles of the triangle. You find the two angles are \begin{align*}45^\circ\end{align*} and \begin{align*}70^\circ\end{align*}. How far are each of the two points on your side of the river from the end of the bridge on the other side of the river? How long will the bridge be? 

7. Draw a picture of this situation.

8. Make a plan: what method(s) or technique(s) can you use to solve this problem?

9. Solve the problem. 

\begin{align*}\Delta ABC\end{align*} has two sides of length 12 and a non-included angle that measures \begin{align*}60^\circ\end{align*}.

10. Draw a possible picture of this situation.

11. Find the measure of all sides and angles of \begin{align*}\Delta ABC\end{align*}.

12. Find the area of \begin{align*}\Delta ABC\end{align*}.

Lily starts at point A and walks straight for 100 feet. Then, she turns right at an \begin{align*}80^\circ\end{align*} angle and continues walking for another 150 feet. In order to go straight back to her starting place, how far will she need to walk? At want angle should she turn right?

13. Draw a picture of this situation.

14. Make a plan: what method(s) or technique(s) can you use to solve this problem?

15. Solve the problem. 

Review (Answers)

To see the Review answers, open this PDF file and look for section 7.8. 

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Vocabulary

Angle of Depression

The angle of depression is the angle formed by a horizontal line and the line of sight down to an object when the image of an object is located beneath the horizontal line.

Angle of Elevation

The angle of elevation is the angle formed by a horizontal line and the line of sight up to an object when the image of an object is located above the horizontal line.

ASA

ASA, angle-side-angle, refers to two known angles in a triangle with one known side between the known angles.

law of cosines

The law of cosines is a rule relating the sides of a triangle to the cosine of one of its angles. The law of cosines states that c^2=a^2+b^2-2ab\cos C, where C is the angle across from side c.

law of sines

The law of sines is a rule applied to triangles stating that the ratio of the sine of an angle to the side opposite that angle is equal to the ratio of the sine of another angle in the triangle to the side opposite that angle.

Tangent

The tangent of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle.

Trigonometric Ratios

Ratios that help us to understand the relationships between sides and angles of right triangles.

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