Impossible Bearing and Triangle word problem.

[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!

 

[eumyang]

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?

 

[amd123]

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.

 

[The legend]

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.

 

[The legend]

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..

 

[zgozvrm]

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!

 

[amd123]

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?

 

[Steed24]

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.

 

[The legend]

dropping perpendicular gave me an easy answer though...

 

[zgozvrm]

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^\circ[/tex]E of point A, therefore \angle[/tex]A must be 90^\circ[/tex] - 54.3^\circ[/tex] = 35.7^\circ[/tex]<br /> Likewise, C lies N35.7^\circ[/tex]W of point B and so \angle[/tex]B must be 54.3^\circ[/tex]<br /> Since \angle[/tex]A + \angle[/tex]B = 90^\circ[/tex] then \angle[/tex]C must be 90^\circ[/tex]<br /> <br /> We have a right triangle.<br /> So the distance BC must be 8.68mi \times[/tex] sin(35.7^\circ[/tex])

 

[The legend]

oh ... foolish me!
i calculated the perpendicular distance from the line joining AB... i should have read the question properly...