I want to know the distance from E to F.

If you draw a perpendicular from AB to XY, and you tell me the distance.

The drawing is not to scale.

Note: Don't tell me that it needs more information, because that's all the information you're gonna get. Seriously, the two missing angles you just don't have them, that's all, forget them.

If it's impossible, tell me, I would be glad to know.

Thanks.

 

This is the JPEG version of the image, sorry, I posted a PDF but I'm sure many of you are not very fond of downloading a file.

 

ASA completely defines a triangle. What more information do YOU need?

John

 

First, I would use the law of sines to find one more side in each triangle.
I would then draw the heights of the triangles that split them in 2, resulting in 4 triangles in total, all with known angles and one known side.
Law of sines again to find the length of the heights and then just add them.

It is very likely that there is a simpler approach to this problem.

 

What more information do YOU need?

What about the distance from E to F?

 

What have you got so far?

Simply giving you the answer won't do much for your education.

We do help people, but it's nice if a bit of initiative is shown.

 

The problem has no unique answer because you need at least one more angle in the top triangle. You've drawn AB and XY so that they appear somewhat parallel, but you didn't state that, so it can't be assumed.

Also, it's not clear what E and F are. Are they points somewhere or did you mean that they are the lines AB and XY, respectively, and that you wanted the length of an interior perpendicular between them? You need to specify your problems more carefully.

If they are parallel, then the corresponding interior angles define the top triangle. Then it's a simple application of the sine law followed by a hypotenuse times a sine to get each triangle's altitude; then add the altitudes together.

Oh, and if they are parallel and you just need a quick answer, make a scale drawing on a piece of paper -- all it takes is a rule, straightedge, and calculator (it's more accurate to construct the angles by using trig than messing with a protractor). I doubt one in a hundred kids would even think of that today, given our wonderful educational systems.

 

If you draw a perpendicular from AB to XY, and you tell me the distance.

This tells me that AB and XY must be parallel. The problem is fully defined.

Maybe this thread should be moved to Homework. The OP should show his efforts at solution.

John

 

I already told you what you should do on post #4. It is not very nice to ignore a legit answer when you don't understand it.

 

#4 will work. He could also use tangents to find the vertical to the bottom triangle. A little algebra is involved, but it is not difficult. Then, because the triangles are identical but of different scale, the ratio of the bases can be used to calculate both heights.

That may be a pedestrian way to do it, but you only need one tangent and a little algebra.

John

 

It is very likely that there is a simpler approach to this problem.

Well I agree with someonesdad that the problem is not clearly posed.

What is a perpendicular from AB to XY?

Strictly it is only guaranteed perpendicular to AB.

If however AB and XY are parallel then the problem is simplified since the triangles are similar so the heights are in the ratio of the known sides.

Note: Don't tell me that it needs more information, because that's all the information you're gonna get. Seriously, the two missing angles you just don't have them, that's all, forget them.

I don't, however, appreciate an attitude like that.

 

So this is really not a math problem, but a grammar problem?

I don't think so. Nor, do I think that refusing to answer based on that most tiny ambiguity will get any points with the teacher. Remember, we have not actually seen the question as written by the teacher, and the OP has not made any attempt to solve the question assuming the most likely interpretation.

Even if the question is poorly written, one will get more points by working it out, showing his approach, and perhaps mentioning the ambiguity than by simply saying it can't be done.

I guess I am a practical guy.

John

 

Well, everything seems to be solved now.

If we pay attention to the old saying that goes:

----"In an isosceles triangle, the angles at the base are equal"

And if we consider AB and XY to be parallel --why not-- then the 2 missing angles of the upper triangle would be 55.5°, which solves the issue.

Right?... I hope so.

Yeah, thanks a lot everyone.

 

No, the top triangle is not isosceles. You need to remember the rule about the angles formed by a line intersecting parallel lines. John

 

Well, everything seems to be solved now.

If we pay attention to the old saying that goes:

----"In an isosceles triangle, the angles at the base are equal"

And if we consider AB and XY to be parallel --why not-- then the 2 missing angles of the upper triangle would be 55.5°, which solves the issue.

Right?... I hope so.

Yeah, thanks a lot everyone.

The angles are not the same though. If the assumption (which must be made) that AB and XY are parallel, the two triangles are similar.

Similar triangles means the angles in the upper triangle are the same as in the lower triangle, but the length of the sides is different (one is a scaled version of the other).

The answer can be found in several ways if AB || XY is assumed.

 

Sorry I hadn't come here for a while. You guys rock, thank you.