# Impossible Bearing and Triangle word problem.

• amd123
In summary, the distance from point B to the nearest hundredth of a mile is 7.62 mi. For the second problem, the car traveled approximately 224.74 feet from when it was first noticed until it stopped.
amd123

## Homework Statement

Radio direction finders are set up at points A and B, 8.68 mi apart on an east-west line. From A it is found that the bearing of a signal from a transmitter is N54.3*E, while from B it is N35.7*W. Find the distance from B, to the nearest hundredth of a mile.

A person is watching a car from the top of a building. The car is traveling on straight road directly toward the building. When first noticed the angle of depression to the car is 28*30’. When the car stops, the angle of depression is 47*8’. The building is 250 feet tall. How far did the car travel from when it was first noticed until it stopped? Round your answer to the hundredths place.

## Homework Equations

All the trigonometric functions

## The Attempt at a Solution

I did these problems on a test and failed at them miserably. I'm trying them again and still can't figure them out. I've set them up and uploaded them:
[PLAIN]http://img684.imageshack.us/img684/7793/setupf.png

Any help is appreciated!

Last edited by a moderator:
Can you show us what work you have? For the first one, have you learned the Law of Sines yet? For the 2nd one, you should know the angles of elevation of the car to the person on top of the building, right? How many right triangles do you see? Which trig function can you set up?

I have not learned law of sines and I know the angles of depression = angles of elevation because of alternate interior angles.

For the bearing problem, I am totally stumped and that's how far I can get.
For the second I really have no idea how to approach it.

ok ... for the second...
see if you have a triangle and a side and angle is given to you can't you find the other by trigonometry? ... ok you can... so from one angle you get the longer dist from tower and from second you get shorter distance... then subtract to get your answer.

for the first drop down a perpendicular 'p' and say the horizontal distance from foot of perpendicular to A is x and other is 8.68-x... apply trig and equate..

The legend said:
for the first drop down a perpendicular 'p' and say the horizontal distance from foot of perpendicular to A is x and other is 8.68-x... apply trig and equate..
No need to drop a perpendicular, if you notice that 54.3 degrees and 35.7 degrees make 90 degrees; you already have a right triangle!

zgozvrm said:
No need to drop a perpendicular, if you notice that 54.3 degrees and 35.7 degrees make 90 degrees; you already have a right triangle!

Yea I knew that, that's obvious. But what do I do after? I can't use a 90* angle to solve trig functions can I?

amd123 said:
Yea I knew that, that's obvious. But what do I do after? I can't use a 90* angle to solve trig functions can I?

Draw a line across the top of the middle triangle in the first problem. This gives you 90* angles in the two triangles that you have now formed on the sides. Fill in the missing angles on each of these triangles. Since there is a straight line across the top, you know that the angles add up to 180*. The top angle of the middle triangle is 90*. Use sin or whatever to solve for the missing side.

dropping perpendicular gave me an easy answer though...

The legend said:
dropping perpendicular gave me an easy answer though...

Your method is fine. It works and it's relatively easy. My point is that the problem is already staring you in the face...

(See the attachment)
Call the angle from the transmitter to points A & B, point C.
We know that C lies N54.3[itex]^\circ[/tex]E of point A, therefore [itex]\angle[/tex]A must be 90[itex]^\circ[/tex] - 54.3[itex]^\circ[/tex] = 35.7[itex]^\circ[/tex]
Likewise, C lies N35.7[itex]^\circ[/tex]W of point B and so [itex]\angle[/tex]B must be 54.3[itex]^\circ[/tex]
Since [itex]\angle[/tex]A + [itex]\angle[/tex]B = 90[itex]^\circ[/tex] then [itex]\angle[/tex]C must be 90[itex]^\circ[/tex]

We have a right triangle.
So the distance BC must be 8.68mi [itex]\times[/tex] sin(35.7[itex]^\circ[/tex])

#### Attachments

• Drawing1-Layout1.png
4.5 KB · Views: 500
oh ... foolish me!
i calculated the perpendicular distance from the line joining AB... i should have read the question properly...

## 1. What is an Impossible Bearing and Triangle word problem?

An Impossible Bearing and Triangle word problem is a type of mathematical problem that involves using bearings and triangulation to find the location of a point or object. It is considered "impossible" because it requires advanced mathematical skills and can be difficult to solve without proper knowledge and techniques.

## 2. How do you approach an Impossible Bearing and Triangle word problem?

To approach an Impossible Bearing and Triangle word problem, you should first understand the basic concepts of bearings and triangulation. Then, carefully read the problem and identify the given information and what you are trying to find. Next, use the appropriate formulas and techniques to solve the problem step by step.

## 3. What are some common mistakes people make when solving Impossible Bearing and Triangle word problems?

One common mistake is not understanding the difference between true bearings and magnetic bearings. Another mistake is not using the correct formulas or not setting up the problem correctly. It is also important to pay attention to units and conversions, as well as rounding errors.

## 4. What are some tips for solving Impossible Bearing and Triangle word problems faster?

Some tips for solving Impossible Bearing and Triangle word problems faster include practicing regularly, memorizing important formulas and concepts, and breaking down the problem into smaller, manageable parts. It can also be helpful to draw diagrams and label given information to visualize the problem better.

## 5. Are there any real-life applications for Impossible Bearing and Triangle word problems?

Yes, Impossible Bearing and Triangle word problems have many real-life applications, particularly in navigation, surveying, and mapping. For example, they can be used to determine the location of an object based on its bearing from multiple points, or to calculate distances and angles between objects in the real world.

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