how to solve bearing word problems

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Bearing - Word Problems

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• Andres Gonzalez
• Marco Faustini
• Mahindra Jain
• Kaustubh Miglani

Most bearing word problems involving trigonometry and angles can be reduced to finding relationships between angles and the measurements of the sides of a triangle . In this case, finding the right basic trigonometric functions to relate the angles and measurements are crucial for setting up and solving the problem correctly.

Trigonometry of angles and sides can be used on a daily basis in the workplace like in carpentry, construction work, engineering, etc.

Basic Definitions

Word problems, example problems.

Before jumping to the word problems, here are the basic definitions you need to be familiar with:

The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer's eye (also known as the line of sight ). The angle of elevation in the above diagram is \(\alpha^\circ\).

If the object is below the level of the observer, then the angle between the horizontal and the observer's line of sight is called the angle of depression . The angle of depression in the above diagram is \(\beta^\circ\).

If the height of a pole is \(2\sqrt3\) metres and the length of its shadow is \(2\) metres, then find the angle of elevation of the sun.

It is often useful to draw a diagram and remember how the basic trigonometric functions relate the angles and measurements of sides in a right triangle. Finding the right trigonometric function to relate the angles and measurements is crucial to solving the problems. We will demonstrate the principle of setting up and solving trigonometric word problems by working through several examples.

A surveyor in a helicopter at an elevation of 1000 meters measures the angle of depression to the far edge of an island as \( 24^\circ \) and the angle of depression to the near edge as \( 31^\circ \). How wide is the island, to the nearest meter? Let the horizontal distance between the helicopter and the island be \( d \), and the width of the island \( w \). Then \( \tan 24^\circ = \frac{1000}{d+w} \) and \( \tan 31^\circ = \frac{1000}{d} \), implying \( d = \frac{1000}{\tan 31^\circ} \). Substituting this into \( \tan 24^\circ = \frac{1000}{d+w} \) gives \[\begin{align} \tan 24^\circ &= \frac{1000}{\frac{1000}{\tan 31^\circ}+w} \\ \tan 24^\circ \left(\frac{1000}{\tan 31^\circ}+w \right) &= 1000 \\ w &= \frac{1000\left( 1 - \frac{\tan 24^\circ}{ \tan 31^\circ }\right)}{\tan 24^\circ} \\\\ & \approx 581.75 . \end{align}\] Thus, the width of the island is 582 meters Alternatively, you can calculate the difference between \(\cot 24^\circ\) and \(\cot 31^\circ\) (which is 0.581757292) and multiply it by 1000 meters. \(_\square\)

Andrew was flying a kite on a hill, but he dumped his kite into the pond below. If the length of the string of his kite is 150 meters and the angle of depression from his position to the kite is \(30^\circ\), then how high is the hill where he is standing? Let's first draw a diagram for a better understanding of the problem: So it is a right triangle with base angle \(30^\circ\), hypotenuse 150 meters, and the side \(h\) opposite to the given angle being the same as the height of the hill. We use the sine ratio to find the height: \[\begin{align} \sin 30^\circ&=\dfrac{h}{150} \\ \dfrac 12 &=\dfrac{h}{150} \\ \\ \Rightarrow h&=75 \text{ (m)}. \end{align}\] Hence the hill is 75 meters above the lake. \(_\square\)

The angle of elevation of the top of an incomplete vertical pillar at a horizontal distance of 100 meters from its base is 45 degrees. If the angle of elevation of the top of the complete pillar from the same spot is to be 60 degrees, then by how much the height of the incomplete pillar should be increased? Let's draw a diagram to figure out the situation: Let \(BC\) be the height of the incomplete pillar, and \(BD\) the height of the complete pillar. We are given that \(BC=100\text{ m}, \angle BAC = 45^\circ,\) and \(\angle BAD = 60^\circ\). And we assume that the length of \(CD\) is \(x\) meters. In triangle \( ABC\), we know that \[\begin{align} \tan 45^\circ & =\frac{BC}{AB} \\\\ AB & =BC \\ & =100 \text{ m}.\end{align}\] Similarly, in triangle \(ABD\), we know that \[\begin{align} \tan 60^\circ & =\frac{BD}{AB} \\ \sqrt 3 & = \frac{BD}{100} \\\\ \Rightarrow BD & =100 \sqrt 3 \text{ m}.\end{align}\] And from the above figure, \[\begin{align} BD&=BC+CD \\ 100\sqrt3 &=100+x \\\\ \Rightarrow x&=100 \big( \sqrt3 -1 \big) \text{ m}. \end{align}\] Hence the height of the incomplete pillar is to be increased by \(100 \big( \sqrt3 -1 \big) \text{ m} \) to complete the pillar. \(_\square\)

Jim and Ted live on one side of the river, and Martha lives on the other side. The distance across the river is 100 yards. Ted, who lives downstream from Matha, measures an angle of 35 degrees between the shoreline and a straight line leading to Martha's house. Jim, who lives upstream from Martha, measures an angle of 60 degrees. How far apart do Ted and Jim live? First off, a picture will help (to get your "bearing"--no pun intended). The relative positions of Martha, Jim, and Ted are represented in this picture: Now, it is much easier to visualize what is going on. The distance in yards from Jim to the position directly across the stream from Matha is given as \[\tan 60^\circ = \frac{100}{x}.\] And the distance in yards from Jim to the position directly across the stream from Matha is given as \[\tan 35^\circ = \frac{100}{y}.\] So, the distance from Ted to Jim is given by \[\begin{align} \text{Distance} &= x + y \\ &= \frac{100}{\tan 60^\circ} + \frac{100}{\tan 35^\circ}\\\\ &= 57.7 + 142.8 \\ &= 200.5 \text{ (yards)}.\ _\square \end{align}\]

A private plane flies for 1.3 hours at 110 mph on a bearing of 40°. Then it turns and continues another 1.5 hours at the same speed, but on a bearing of 130°. At the end of this time, how far is the plane from its starting point? The bearings tell us the angles from "due north" in a clockwise direction. Since 130 – 40 = 90, the two bearings give us a right triangle. From the times and rates, we have \[\begin{align} 1.3 × 110 &= 143\\ 1.5 × 110 &= 165. \end{align}\] Now, let's give the geometrical shape to our problem and set up a triangle: Using the Pythagorean theorem , we get \[\begin{align} m^2 & =143^2 + 165^2 \\ & =20449 + 27225 \\ & =47674 \\\\ \Rightarrow m & = 218.34. \end{align} \] Hence the plane is approximately 218 miles away at the end of the time. \(_\square\)

Try the following bearing word problems:

A pole of 8-feet height is located on top of a house, right on the edge of the ceiling. From a point on the ground, the angle of elevation to the top of the house is \(17^\circ\) and the angle of elevation to the top of the pole is \(21.8^\circ.\)

Find the height of the house.

A tower stands at the center of a circular park. \(A\) and \(B\) are two points on the boundary of the park such that \(AB\) subtends an angle of \( 60^\circ\) at the foot of the tower and the angle of elevation of the top of the tower from \(A\) or \(B\) is \( 30^\circ.\)

Find the height of the tower.

This section is meant to enhance the problem-solving skills of bearing word problems. Here an example is illustrated followed by some problems for you to attempt:

A six-meter-long ladder leans against a building. If the ladder makes an angle of \(60^\circ\) with the ground, (1) how far up the wall does the ladder reach, and (2) how far from the wall is the base of the ladder? To understand the situation, let's draw a diagram: (1) From the above diagram, we get the following equation to obtain the value of \(h:\) \[\begin{align} \sin 60^\circ & =\dfrac h6 \\ \Rightarrow h & =6 \times \sin 60^\circ \\ & =3 \sqrt 3. \end{align} \] So the ladder can reach \(3\sqrt3 \) meters up the wall. \[\] (2) Similarly, we can get the value of \(b\) as follows: \[\begin{align} \cos 60^\circ &=\dfrac b6 \\ \Rightarrow b&=6 \times \cos 60^\circ \\ &=3. \end{align}\] Hence the base of the ladder is 3 meters from the wall. \(_\square\)

Here are the problems to gain a strong grab over the bearing concept:

The angle of elevation of a cloud from a point \(h\) meters above a lake is \(\theta\). The angle of depression of its reflection in the lake is \(45^{\circ}\). Find the height of the cloud in meters.

Linda measures the angle of elevation from a point on the ground to the top of a tree and finds it to be 35 degrees. She then walks 20 meters towards the tree and finds the angle of elevation from this new point to the top of the tree to be 45 degrees. Find the height of the tree (in meters).

Give your answer to three significant figures.

From the top of a 7-meter-high building, the angle of elevation of the top of a cable tower is \(60°\) and the angle of depression of its foot is \(45°.\) Determine the height of the tower in meters.

If the angles of elevation of the top of a tower from three collinear points \(A, B,\) and \(C\), on a line leading to the foot of the tower, are \(30^{\circ},45^{\circ},\) and \(60^{\circ},\) respectively, then the ratio \(AB:BC\) is \(\text{__________}.\)

Lengths in Right Triangles

Trigonometric Equations

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Here we will learn about bearings, including measuring bearings, drawing bearings and calculating bearings.

There are also bearings maths worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What are bearings?

Bearings are angles, measured clockwise from north.

To measure a bearing, we must first know which direction is north.

This north direction is usually provided in the maths exam question. We then measure the required angle in a clockwise direction. All bearings need to be given in three figures, so if the angle measured is less than 100 degrees, we must start the three-figure bearing with a zero.

The diagram shows three points A, B and P.

The angles are measured clockwise from the north line. 

The bearing of A from P is 045^{\circ} . 

The bearing of B from P is 260^{\circ} .

Bearings are used by sailors and pilots to describe the direction they are travelling. They are also used on land by hikers and the military.

How to draw bearings

In order to draw bearings:

Locate the point you are measuring the bearing from and draw a north line if there is not already one given.

Using your protractor, place the zero of the scale on the north line and measure the required angle clockwise, make a mark on your page at the angle needed.

Draw a line from the start point in the direction of the bearing. If you are producing a scale drawing and know the distance to locate a point use this scale appropriately.

Explain how to draw bearings

Bearings maths worksheet

Get your free bearings maths worksheet of 20+ questions and answers. Includes reasoning and applied questions.

Related lessons on loci and constructions

Bearings  is part of our series of lessons to support revision on  loci and construction . You may find it helpful to start with the main loci and construction lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:

• Loci and construction
• Constructions
• Perpendicular bisector
• Angle bisector
• Constructing triangles
• How to draw a hexagon
• How to construct a 30, 60, 45, 90 degree angle
• Constructions between points and lines

Bearings maths examples (drawing)

Example 1: drawing a bearing less than 180°.

Draw a bearing of 050^{\circ}

2 Using your protractor, place the zero of the scale on the north line and measure the required angle clockwise, make a mark on your page at the angle needed.

3 Draw a line from the start point in the direction of the bearing. If you are producing a scale drawing and know the distance to locate a point use this scale appropriately.

Example 2: drawing a bearing more than 180°

Draw a bearing of 300^{\circ}

If you have a 180^{\circ} protractor, we need to subtract the bearing required from 360^{\circ}

360^{\circ} - 300^{\circ} = 60^{\circ} , so measure 60^{\circ} anticlockwise

Example 3: drawing a scale drawing with a bearing

Make a scale drawing of a point Q 8 km away from a point P on a bearing of 110^{\circ} from P, using a scale of 1 cm : 2 km.

If the angle is more than 180^{\circ} and you do not have a full circle protractor then you should subtract the required angle from 360^{\circ} , and then measure this angle anticlockwise from the north line.

If the scale is 1cm : 2km , we will need to measure 4 cm to locate Q.

How to calculate bearings

In order to calculate bearings:

Locate the points you are calculating the bearing from and to.

Using the north lines for reference at both points, use angle rules and/or trigonometry to calculate any angles that are required.

Read off the three-figure bearing required.

Bearings maths examples (calculating)

Example 1: calculating a bearing around a point.

Calculate the bearing of A from P.

We need the angle around the point P clockwise from the north line. We know that angles on a straight line add to 180^{\circ}. 180^{\circ} + 53^{\circ} = 233^{\circ}

The bearing of A from P is 233^{\circ} .

Example 2: calculating a back bearing

The bearing of B from A is 070^{\circ} . Calculate the bearing of A from B.

Drawing a north line at B and extending the line from A to B shows us a corresponding angle. If we travel from A to B, we need to turn another 180^{\circ} to return back to A. 180^{\circ} + 70^{\circ} = 250^{\circ}

The bearing of A from B is 250^{\circ} .

Example 3: calculate a bearing using SOHCAHTOA

A ship sails 7 km due east from a point P to point A. It then sails 3 km due south from A to point B. Calculate the bearing of B from P.

Sketch the diagram to help visualise the problem.

Drawing a line from P to B forms a right angled triangle. Use trigonometry to find the angle APB. Then add this to 90^{\circ} .

{{\tan }^{-1}}\left( \frac{3}{7} \right)=23^{\circ} to the nearest degree (bearings are normally given to the nearest degree)

90^{\circ} +23{}^\circ =113^{\circ}

The bearing of B from P is 113^{\circ} .

Common misconceptions

• Not giving bearings less than 100^{\circ} in three figures

E.g. A bearing with an angle of 45^{\circ} must be given as 045^{\circ} .

• Using the anticlockwise angle as the bearing

A common error is to use the anticlockwise angle as the bearing. E.g. The bearing of B from A is 080^{\circ} , a common mistake would be to use the co-interior angle as the back bearing of A from B as 100^{\circ} , calculating clockwise, the bearing of A from B should be 260^{\circ} .

Practice bearings maths questions

1. Which of the diagrams shows a bearing of 040^{\circ} ?

Angle should be measured clockwise from north.

2. Which of the diagrams shows a bearing of 290^{\circ} ?

3. What bearing describes due east?

north is 000^{\circ} , east is 090^{\circ} , west is 270^{\circ} and south is 180^{\circ} .

4. Calculate the bearing of A from P.

Angle should be measured clockwise from north, so subtract 125^{\circ} from 360^{\circ} .

5. The bearing of C from B is 130^{\circ} . Calculate the bearing of B from C.

6. A boat sails 8 km north from P to Q and then sails 6 km west from Q to R. Calculate the bearing of R from P. Give your answer to the nearest degree.

Sketch a diagram. Angle should be measured clockwise from north, so find the angle QPR using trigonometry and subtract from 360^{\circ} .

Bearings maths GCSE questions

1. The bearing of A from B is 215^{\circ} . Find the bearing of B from A.

Either 215 – 180 or 360 – 215 = 145 seen

2. The point C is on a bearing of 065^{\circ} from point A and on a bearing of 310^{\circ} from point B.

On the diagram, mark with a (x) the position of point C.

Correct line from A or correct line from B

Both lines correct and (x) shown

3. The diagram shows the positions of three twins labeled P, Q and R.

Q is on a bearing of 080^{\circ} from P.

R is on a bearing of 132^{\circ} from P.

The distance PQ is 15 km and the distance PR is 14 km.

a) Find the distance QR

b) Find the bearing of R from Q

Angle QPR = 132^{\circ}- 80^{\circ}= 52^{\circ}

Values substituted into cosine rule QR ^2=15^2+14^2-2 \times 15 \times 14 \times \cos 52

QR = 12.74 km

Use of sine rule \frac{\sin{52}}{12.74}=\frac{\sin{PQR}}{14}

Angle PQR = 59.96^{\circ}

Angle of 100^{\circ} anticlockwise from north line at Q

Answer of 200^{\circ}

Learning checklist

You have now learned how to:

• Measure line segments and angles in geometric figures, including interpreting maps and scale drawings and use of bearings

The next lessons are

• Angles in parallel lines
• Trigonometry
• Cosine rule

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Tips on Solving Word Problem with Bearings

All right, guys, in this example, I want to cover some tips to help you with this word problem with bearings which usually confuses the heck out of students.  So hopefully, some of the tips that I’m going to cover in this video will help you understand how to solve a word problem and give you a better understanding of how to use and solve for bearings.

Part One: The Problem 

So in this example, we have a ship that travels 80 miles east from Port.  Now, I’m just going to stop.  Right.  Because one of the first things that we always want to do when we have a word problem is to draw a picture.  And you’ve probably heard from your teacher before.  When you’re doing word problems, draw a picture, draw a picture, draw a picture.  Right.  The reason why that’s so helpful is it allows us to visualize what exactly is happening because otherwise, you’re just looking at these words, and it’s like blah.  Right.  So our primary goal is extracting a picture, extracting mathematical equations that we can then go ahead and understand and solve for it.  So I have a ship that travels 80 miles to ease.  So whenever I have this kind of, like, bearing and direction, the first thing we’re going to do is create a Cardinal direction.

Then from there, what you’re going to do is you’re going to have the north right, east, south, and west.  Now we have this point here.  We’re going to call that go ahead and P for our Port.  Now, it says the ship travels 80 miles to the east.  So I need to go ahead and represent that somehow.  So what I’m going to do is go ahead and draw a nice little line going out here.  And then we can go ahead and end and let’s go ahead and label that to 80 miles.  We’ll leave the units off for a moment.

Then it says it takes a turn in the northeast direction.  Now, the northeast is very vague.  We don’t know where it’s going.  We know it’s in this direction.  Somewhere over in here, it says you’re going to go in that direction for another 60 miles.  So again, we have this northeast direction.  So whenever I have this, I need to go ahead and create another Cardinal direction.  So again, from this turning point here, I will go ahead and do exactly everything over.  Again, I’m going to do north, and I’m just going to label this east, south, and west.  Now, again, everything doesn’t need to look perfect.  And everything else, hopefully, you’re getting the idea.  Now we’re going in this direction.  We’re going 60 miles.  So again, I’m just going to go ahead and represent this with another line here.  And nothing has to be perfect.  Ladies and gentlemen, let’s go ahead and say here are going to be 60 miles.  And then it says if the boat is now 130 miles away from Port.  So now I’m going to connect the Port to where this boat is here.  And I’m going to say that that distance is 139 miles.

Now it’s asking us what is the bearing from Port to the ship and what is the bearing from the ship back to Port?  So here’s where the ship is currently.  Here’s where its Port is, right.  It traveled 80 miles to the east, then took a turn right out to this direction.  When we’re looking for the bearing, what we’re looking for is we’re looking for an angle from due north.  So there are a couple of different ways we can write bearings.

We’re looking for this angle, and we have a problem here because we don’t have a way to solve this angle outside of the triangle.  However, hopefully, you’ll recognize here I have an oblique triangle and this oblique triangle.  I have all three side lengths.  Therefore, we have a way to solve this triangle, which is going to be using the law of cosines.  Another tip that I like to kind of recommendations to my students.  When you’re looking at a problem like this, there’s already a lot going on.  If you look at the words, sometimes it’s like mental overload.  Now you create a picture to represent and extract some of the information.  Now it’s also a kind of cognitive overload.  I know what I need to do.  I need to find the angle right here inside this triangle.

So to go ahead and do that, what I’m simply going to do is I’m going to redraw this one more time.  Still, I’m just going to focus on this mathematically, and I’m going to rewrite them using A, B, and C because that’s how I like to solve my oblique triangles.  So what I’m going to do is I’m just going to redraw the triangle.  And again, it can look very similar to the one I just did.  But forgetting about the Port and the ship and all that kind of stuff, I’m just going to label my oblique triangle the way I like to label all my oblique triangles, just like this.  Remember, we have our opposing side lengths.  A this is going to be B, and this is going to be C.  From this information, if we need to go ahead and solve for A, I know I can use the law of cosine here.

So hopefully, you either have this memorized, or it’s on the board, or your teacher gives you a formula sheet that you can go ahead and reference.  Okay.  Now, it doesn’t matter what angles we’re dealing with.  We can all interchange them.  A, B, and C.  This will be your typical formula for the law of cosines when we want to solve for an angle.  Now, the cool thing is we know what B is.  That’s going to be 80.  We know what A is.  That’s 60.  And we know C, which is 139.  So now, I’m going to go ahead and plug them into my formula.  Okay.  And now, to go ahead and get to this next point, I’m going to want to use my calculator.  Right.  So remember, when I want to solve for my angle A, we’re not solving for the cosine of A; we’re solving for the angle of A.  So make sure that you undo the cosine by now.  Go ahead and type in the cosine inverse of all this information.  Okay.  And now I’m just going to plug this into my calculator.  Just make sure you’re careful when plugging it in, and then make sure you also take the cosine inverse.  So, therefore, we can find our angle A.

So now you can see that A is equal to five-point 92 degrees, which is a pretty slight angle.  And if you look at this triangle, like initially how I had it, that can make some sense.  And again, that’s important about drawing your angles because, if this angle is like obtuse or something, you say something seems off.  So that’s why it’s always nice to have a general picture of your triangle from the original problem.  So you can make sense of if your answer is correct.  Remember, that’s the angle right here.  We’re looking for this angle.  So the nice thing is, remember, these two angles will be complementary.  So these angles are going to be complementary.

So all I need to do to find this angle is subtract this angle from 180 degrees.  And again, this is the rounded version of the angle.  But again, inside my calculator, I’m going to use the exact answer, and therefore, I get a rise of 84.07 degrees.  When I want to go and answer this question, I’ll say the bearing from Port to the ship is 84.07 degrees.  Or another way, we could say that it will be north 84.07 degrees east.  So that’s just another way we could do bearings.

Part Two: The Problem

The next question we have here is finding the bearing from where the ship is back to Port.

So you can think about the first answer: let’s say you have station control at Port, right?  They’re trying to see the bearing from the Port to the ship.  Let’s say they needed to go ahead and find them.  That’s why that would be important.  Now, the other answer is from the ship to the Port.  Let’s say they need to return to Port.  And what direction do they need to turn?  So, to find that answer, we need to create another Cardinal direction.  So what I want you to recognize here is if this is going to be my due North Mrs. East southwest, again, my bearings will go from due north.  So what I need to do is find the angle from here.  Now that’s going to be very difficult to go ahead and do because how do I feel to see this angle?  The only thing I can figure out is how to find this angle.  And what you could do is also extend this line right here.  And if you think about parallel lines in the transversal, remember we have those angle relationships, corresponding angles, and alternate interior angles.

So the one thing I want you to recognize here is that angle and that angle is going to be equal to each other.  So, therefore, if I can find this angle and that angle, I’m going to be able to cover that side, and then I need to add 90 degrees.  The nice thing about having a triangle is I only need to figure out one more angle here to go ahead and add these two to find my missing third side. So going back to my triangle, now I’m going to want to go ahead and find this angle C and then go ahead and find my angle B.  But again, I can subtract those two from there.  Now it’s very important to make sure that you stored answer A, so, therefore, you have that in your calculator because, again, remember, we’re going to want to subtract that from 180.

Now, we will write the law of cosines to find angle C.  So remember, A will be 60, B is equal to 80, and C equals 139.  Again, remember we’re trying to solve for C. Therefore, I will take the cosine inverse to undo the cosine.  So I’m going to plug this all into my calculator that C equals the cosine inverse.  Okay, when I do that, I get 166.15, which will be seen.  And again, I’m going to store this because we already had A, which I wanted to store.  And now what we need to be able to find is going to be B.  So again, remember if I have two angles of a triangle, I can subtract those from 180 to find my missing angle.  So B will be 180 degrees minus the stored answer for A, minus the stored answer for C.

When I do that again using my calculator and storing this, I’m going to go and have 7.92 degrees.  Again, that is going to be my answer B.  Now again, we have to remember what all this context means.  Why is this again helping me find my bearing?  So what I want to do is go back to the original picture so we can go and take a look.

So again, remember what we found?  We found this angle C because of alternate interior angles; this line is parallel.  They both intersect with this line.  So C is going to be equivalent to this angle as well.  From here, that’s going to be from this red line to the X-axis.  Then we also just found B, which will be this angle as well.  So if I add B plus C plus 90 degrees, that’s going to give me the bearing from due north over to this angle because you’re figuring a ship at this point, and you need to find the bearing back to the Port we’re going to go from due north all the way around to this line which is going to take you back to Port.  So, therefore, I’m just going to take my angle B plus my angle C, and I’m going to add 90 degrees.  That is going to give me my bearing.  So when I do that, I get 264.7 degrees.  Now again, let’s go ahead and check our work. Because again, remember we created this original graph here, so if I was to go ahead and grab this from due north 264 degrees.  Well, from here to here is 90.  From here to here is 180, and from here to here is 270.  So 264 will be just a little bit shorter, right?  So it’s going to look something like that probably.  Does that look like the angle I need to go from the ship back to the Port?  Yeah, not too bad, right?  You can see I’m in the right frame.  It’s not perfect.  My drawing is not pictured but again, and you can see that this angle, this bearing here, is roughly a pretty good match going back to my Port.  Okay, so now I need to finalize again.  The answer did a lot of work, so we’re going to say the bearing from the ship to Port is 260. 64.7 degrees.

So, there you go, ladies and gentlemen.  Hopefully, this was helpful for you and some of the tips that I have for you along the way.  If you want more examples, check out the playlist and resources I have for you below, or check out the following video .

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Trigonometry : Bearing

Study concepts, example questions & explanations for trigonometry, all trigonometry resources, example questions, example question #1 : bearing.

Example Question #2 : Bearing

Example Question #3 : Bearing

The following diagram could represent which one of these practical scenarios?

Example Question #4 : Bearing

17.5 miles south and 30.31 miles east

30.31 miles south and 17.5 miles east

105 miles south and 181.87 miles east

181.87 miles south and 105 miles east

First, let's set up a diagram using the given information. This looks like this:

Now we can use trigonometry to determine the missing sides, s and e.

Therefore the ship has travelled 181.87 miles south and 105 miles east. 

Example Question #5 : Bearing

The airplane is 139.46 km south of its starting point and 299.08 km east of its starting point.

The airplane is 139.46 km south of its starting point and 707.69 km east of its starting point.

The airplane is 707.69 km south of its starting point and 139.46 km east of its starting point.

The airplane is 299.08 km south of its starting point and 139.46 km east of its starting point.

The question is asking us how far south and how far east the plane is from its starting point, so we need to now use trigonometry to determine the lengths of the missing sides of the triangle. We will call these sides s for the southward distance and e for the eastward distance.

Therefore the airplane is 139.46 km south of its starting point and 299.08 km east of its starting point.

Distance from Ship A to Ship B: 640.67 miles

Distance from Ship B to Ship C: 594.02 miles

Distance from Ship A to Ship B: 594.02 miles

Distance from Ship B to Ship C: 640.67 miles

Next, we need to use trigonometry to find the answers to each question we're being asked. To find the distance AB, set up

Next, we can find the distance BC. There are two ways to do this since we know two angles of the triangle, but either way you need to use the tangent function.

Example Question #22 : Practical Applications

Three cruise ships are situated as follows: Sea Terraformer is 200 miles due north of Wave Catcher, and Island Pioneer is 345 miles due east of Wave Catcher. What is the bearing from Island Pioneer to Sea Terraformer, and what is the bearing from Sea Terraformer to Island Pioneer?

Begin by diagramming the given information; you'll see that the three ships create a right triangle. To solve the question, we need to find the unknown angles of the triangle, then frame our answers as the proper bearings. 

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Bearing Problems & Navigation

• > Trigonometry
• > Bearings

Bearings and direction word problems

• Examples 0/8

Get the most by viewing this topic in your current grade. Pick your course now .

• Introduction to Bearings and Direction Word Problems

Charlie leaves home for a bike ride, heading 040°T for 5km.

• How far north or south is Charlie from its starting point?
• How far east or west is Charlie from its starting point?

A camping group made a return journey from their base camp. From the camp, they first travelled 120°T for 3km. Then they travelled 210°T for 9km. Determine the direction and distance they need to travel if they want to return to the base camp now.

Melody and April go to the same school. Melody's home is 3.5km with a bearing of S16°W from school whilst April's home is 2.4km with a bearing of N42°E from school. How far away are their homes from each other?

Radar X detected an earthquake N55°E of it. 16km due east of Radar X, Radar Y detected the same earthquake N14°W of it.

• Determine the earthquake from Radar X and Y.
• Which Radar is closer to the earthquake?

A plane is sighted by Tom and Mary at bearings 028°T and 012°T respectively. If they are 2km away from each other, how high is the plane?

Consider the following diagram.

Find the distance between P and Q.

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Topic Notes

Theorems that are useful:

Pythagorean Theorem: a 2 + b 2 = c 2 a^{2} + b^{2} = c^{2} a 2 + b 2 = c 2

Trig ratio: sin ⁡ θ = O H \sin \theta = \frac{O}{H} sin θ = H O ​

cos ⁡ θ = A H \cos \theta = \frac{A}{H} cos θ = H A ​

tan ⁡ θ = O A \tan \theta = \frac{O}{A} tan θ = A O ​

Law of sine: a sin ⁡ A = b sin ⁡ B = c sin ⁡ C \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} s i n A a ​ = s i n B b ​ = s i n C c ​

Law of cosine: c 2 = a 2 + b 2 − 2 a b cos ⁡ C c^{2} = a^{2} + b^{2} - 2ab \cos C c 2 = a 2 + b 2 − 2 ab cos C

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Solve Navigation Problems using Vectors

Examples, solutions, videos, worksheets, games and activities to help PreCalculus students learn to solve navigation problems using vectors.

Related topics: More PreCalculus Lessons More Lessons on Vectors

Navigation Problems The word problems encountered most often with vectors are navigation problems. These navigation problems use variables like speed and direction to form vectors for computation. Some navigation problems ask us to find the groundspeed of an aircraft using the combined forces of the wind and the aircraft. For these problems it is important to understand the resultant of two forces and the components of force.

How we measure airspeed, wind, groundspeed, and other variables in navigation?

How to solve application problems using vectors? This video explains 2 basic application problems involving vectors. Example 1: A ship leaves port on a bearing of 28° and travels 7.5 miles. The ship then turns due east and travels 4.1 miles. How far is the ship from the port and what is its bearing?

Example 2: Two tow trucks are pulling on a truck stuck in the mud. Tow truck #1 is pulling with a force of 635 lbs at 51° from the horizontal while tow truck #2 is pulling with a force of 592 lbs at 39° from the horizontal. What is the magnitude and direction of the resultant force?

Word Problems Involving Velocity or Other Forces (Vectors) In this problem we do a word problem involving the bearing (direction) of a boat. Example: A boat is traveling at a speed of 30 mph. The vector that represents the velocity is 15<√2, -√2>. What is the bearing of the boat?

Word Problems Involving Velocity or Other Forces (Vectors) In this problem we are given the bearing and velocity of a plane and the bearing and velocity of the wind; we want to find out the actual velocity of the plane after taking the wind into consideration Example: A plane leaves the airport on a bearing 45° traveling a 400 mph. The wind is blowing at a bearing of 135° at a speed of 40 mph. What is the actual velocity of the plane?

Word Problems Involving Velocity or Other Forces (Vectors) In this problem we are given the force required to keep a box from sliding down a ramp. We want to know the level of inclination of the ramp. Example: A 700 lb force just keeps a 4000 lb box from sliding down a ramp. What is the angle of inclination of the ramp?

Airplane and Wind Vector Word Problems Example: An airplane is flying in the direction 15° North of East at 550 mph. A wind is blowing in the direction 15° South of East at 45 mph. a) Find the component form of the velocity of the airplane and the wind. b) Find the actual speed (“ground speed”) and direction of the airplane.

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How do you solve word problems in math?

Master word problems with eight simple steps from a math tutor!

Author Amber Watkins

Published April 2024

• Key takeaways
• Students who struggle with reading, tend to struggle with understanding and solving word problems. So the best way to solve word problems in math is to become a better reader!
• Mastery of word problems relies on your child’s knowledge of keywords for word problems in math and knowing what to do with them.
• There are 8 simple steps each child can use to solve word problems- let’s go over these together.

Table of contents

• How to solve word problems

Lesson credits

As a tutor who has seen countless math worksheets in almost every grade – I’ll tell you this: every child is going to encounter word problems in math. The key to mastery lies in how you solve them! So then, how do you solve word problems in math?

In this guide, I’ll share eight steps to solving word problems in math.

How to solve word problems in math in 8 steps

Step 1: read the word problem aloud.

For a child to understand a word problem, it needs to be read with accuracy and fluency! That is why, when I tutor children with word problems, I always emphasize the importance of reading properly.

Mastering step 1 looks like this:

• Allow your child to read the word problem aloud to you. 
• Don’t let your child skip over or mispronounce any words. 
• If necessary, model how to read the word problem, then allow your child to read it again. Only after the word problem is read accurately, should you move on to step 2.

Step 2: Highlight the keywords in the word problem

The keywords for word problems in math indicate what math action should be taken. Teach your child to highlight or underline the keywords in every word problem. 

Here are some of the most common keywords in math word problems: 

• Subtraction words – less than, minus, take away
• Addition words – more than, altogether, plus, perimeter
• Multiplication words – Each, per person, per item, times, area 
• Division words – divided by, into
• Total words – in all, total, altogether

Let’s practice. Read the following word problem with your child and help them highlight or underline the main keyword, then decide which math action should be taken.

Michael has ten baseball cards. James has four baseball cards less than Michael. How many total baseball cards does James have? 

The words “less than” are the keywords and they tell us to use subtraction .

Step 3: Make math symbols above keywords to decode the word problem

As I help students with word problems, I write math symbols and numbers above the keywords. This helps them to understand what the word problem is asking.

Let’s practice. Observe what I write over the keywords in the following word problem and think about how you would create a math sentence using them:

Step 4: Create a math sentence to represent the word problem

Using the previous example, let’s write a math sentence. Looking at the math symbols and numbers written above the word problem, our math sentence should be: 10 – 5 = 5 ! 

Each time you practice a word problem with your child, highlight keywords and write the math symbols above them. Then have your child create a math sentence to solve. 

Step 5: Draw a picture to help illustrate the word problem

Pictures can be very helpful for problems that are more difficult to understand. They also are extremely helpful when the word problem involves calculating time , comparing fractions , or measurements . 

Step 6: Always show your work

Help your child get into the habit of always showing their work. As a tutor, I’ve found many reasons why having students show their work is helpful:

• By showing their work, they are writing the math steps repeatedly, which aids in memory
• If they make any mistakes they can track where they happened
• Their teacher can assess how much they understand by reviewing their work
• They can participate in class discussions about their work

Step 7: When solving word problems, make sure there is always a word in your answer!

If the word problem asks: How many peaches did Lisa buy? Your child’s answer should be: Lisa bought 10 peaches .

If the word problem asks: How far did Kyle run? Your child’s answer should be: Kyle ran 20 miles .

So how do you solve a word problem in math?

Together we reviewed the eight simple steps to solve word problems. These steps included identifying keywords for word problems in math, drawing pictures, and learning to explain our answers. 

Is your child ready to put these new skills to the test? Check out the best math app for some fun math word problem practice.

Parents, sign up for a DoodleMath subscription and see your child become a math wizard!

Amber Watkins

Amber is an education specialist with a degree in Early Childhood Education. She has over 12 years of experience teaching and tutoring elementary through college level math. "Knowing that my work in math education makes such an impact leaves me with an indescribable feeling of pride and joy!"

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How to Solve These Basic Trigonometry Questions (Bearings, Word Problems)?

Hi all, i know you are all busy, but i don't understand how to solve these, please help please fully answer all parts of the question, thank you question #5 question #18 and # 19 question #3 and # 4.

The given triangle is isosceles in which the equal sides are #2/3# of the base. So let us consider an isosceles triangle ABC where the base BC is 6 unit and the equal sides are 4 unit . The base angle is # theta# . AD is perpendicular from A to BC.

It is obvious from figure that #costheta = "adjacent"/"hypotenuse" =3/4#

So #theta = cos^-1(3/4)=41.4^@#

Question 19

As per given condition of the question the second isosceles triangle (EBC) has same base that of first one (ABC) but the area of second one is thrice that of first one. It is possible only if the height of second triangle is thrice that of first one. Since the area of the triangle is proportional to height when base is constant.

This has been shown in the fig below.

The perpendicular drawn from vertex of an isosceles triangle bisects the base.

#(DeltaEBC)/(DeltaABC)=(1/2xxBCxxED)/(1/2xxBCxxAD) #

#=>3=(ED)/(AD)#

Now #(tan/_ECB)/(tan/_ACB)=((ED)/(BC))/((AD)/(BC))=(ED)/(AD)#

#=>(tantheta/tan24^@ )=3#

#=>tantheta=3xxtan24^@=1.34#

#=>theta =tan^-1(1.34)~~53.2^@#

Question 3a

I) #B" from "A->41^@#

II) #C" from "B->142^@#

III) #B" from "C->(279+43)^@=322^@#

IV) #C" from "A->(41+58)^@=99^@#

V) #A" from "B->(142+38+41)^@=221^@#

VI) #A" from "C->279^@#

Question 3b

I) #B" from "A->27^@#

II) #C" from "B->151^@#

III) #B" from "C->(246+85)^@=331^@#

IV) #C" from "A->(27+39)^@=66^@#

V) #A" from "B->(151+29+27)^@=207^@#

VI) #A" from "C->246^@#

Bearing = # 90^@ +tan^-1(9/14)~~90^@+33^@=123^@#

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• How do you solve word problems?
• To solve word problems start by reading the problem carefully and understanding what it's asking. Try underlining or highlighting key information, such as numbers and key words that indicate what operation is needed to perform. Translate the problem into mathematical expressions or equations, and use the information and equations generated to solve for the answer.
• How do you identify word problems in math?
• Word problems in math can be identified by the use of language that describes a situation or scenario. Word problems often use words and phrases which indicate that performing calculations is needed to find a solution. Additionally, word problems will often include specific information such as numbers, measurements, and units that needed to be used to solve the problem.
• Is there a calculator that can solve word problems?
• Symbolab is the best calculator for solving a wide range of word problems, including age problems, distance problems, cost problems, investments problems, number problems, and percent problems.
• What is an age problem?
• An age problem is a type of word problem in math that involves calculating the age of one or more people at a specific point in time. These problems often use phrases such as 'x years ago,' 'in y years,' or 'y years later,' which indicate that the problem is related to time and age.

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• Middle School Math Solutions – Inequalities Calculator Next up in our Getting Started maths solutions series is help with another middle school algebra topic - solving...

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