Drawing a Triangle with Side a=4 cm, Altitude 3 cm, and Angle α=60°

[skrat]

Homework Statement

Side ##a=4 cm##, altitude to side a is 3 cm , angle ##\alpha =60 °##.

How can I draw that? Step by step

Homework Equations

The Attempt at a Solution

 

[Mentallic]

What are you allowed to use in your construction? A ruler and protractor for example would make this a trivial task.

Also, have you tried anything?

 

[skrat]

Am... Trivial?

:D

I tried:
Decided where my point A is. :D
Than using a thumbscrew compass I measured the altitude to side a and drew a circle around point A.
I just picked a random radius of that circle, and drew a perpendicular line to it. (on this line, my a side should be).
However, the problem with this is I can't find a way to exactly determine where the other two points are now...

 

[Mentallic]

Ok hang on, just so I have a clear understanding of what your triangle dimensions are.

If we draw side a=4cm alone the bottom, let A be the left point, B the right point (hence AB is 4cm), and C the top point that will connect the triangle together. Now which length is 3cm long and which angle is 60^o?

 

[skrat]

If we draw side a=4cm alone the bottom, let A be the left point, B the right point (hence AB is 4cm)...

I didn't read any further.

I use this notation: http://en.wikipedia.org/wiki/Triangle#mediaviewer/File:Triangle_with_notations_2.svg

So if you draw side a, you have points B and C.

 

[pbuk]

If angle ## \alpha ## is adjacent to side a it is easy, but I assume this is not the case.

So you have two right angled triangles one with an angle of ## \theta ## adjacent to a side of length 3 cm and opposite a side of length ## x ##, and one with an angle of ## 60 - \theta ## adjacent to a side of length 3 cm and opposite a side of length ## 4 - x ## cm.

Two equations, two unknowns - and a very surprising (to me) solution.

Not sure if you can do this with ruler, protractor and compass though.

 

[Mentallic]

You'll be surprised with how much time is spent wasted because the problem was misunderstood and the standards of triangle labelling aren't always upheld.

You can use the sine and cosine rule to find length c. If b = 3cm then with the cosine rule

[tex]c^2 = a^2+b^2-2ab\cdot \cos C[/tex]

Plugging our known values in

[tex]c^2=3^2+4^2-2\cdot 3\cdot 4\cos C[/tex]

[tex]c^2=25-24\cos C[/tex]

Now for the sine rule

[tex]\frac{\sin A}{a} = \frac{\sin B}{b}[/tex]

Plugging values

[tex]\sin B = \frac{3\sqrt{3}}{8}[/tex]

Hence you've got B and then you can find [itex]\sin C[/itex]. I'm assuming you're only using a compass, because if you could use a protractor then just the sine rule would be sufficient to solve your problem.

 

[skrat]

You'll be surprised with how much time is spent wasted because the problem was misunderstood and the standards of triangle labelling aren't always upheld.

Not really. :D
I understood it exactly the same way you did. But I later found out that it is not so simple. I am sorry for being evil.

You can use the sine and cosine rule to find length c. If b = 3cm then with the cosine rule

[tex]c^2 = a^2+b^2-2ab\cdot \cos C[/tex]

Plugging our known values in

[tex]c^2=3^2+4^2-2\cdot 3\cdot 4\cos C[/tex]

[tex]c^2=25-24\cos C[/tex]

Now for the sine rule

[tex]\frac{\sin A}{a} = \frac{\sin B}{b}[/tex]

Plugging values

[tex]\sin B = \frac{3\sqrt{3}}{8}[/tex]

Hence you've got B and then you can find [itex]\sin C[/itex]. I'm assuming you're only using a compass, because if you could use a protractor then just the sine rule would be sufficient to solve your problem.

That works perfectly!
Thank you!

 

[skrat]

Ok, or maybe it doesn't work perfectly.

Who said b=3cm?
If you used b just for labelling, than I am sorry to say that in that case the value of a is no longer 4 cm.

Or is there something I missunderstood?

 

[Mentallic]

EDIT: incorrect post ahead.


Ugh... altitude = 3cm... I took that as being the side length of the triangle. Sorry, it's been a long night. That's another thing that'll waste everyone's time

Just once more... Altitude to side a means that the altitude of 3cm connects point A to side a perpendicularly, correct?

If that is the case (which at this point I wouldn't bet heavily on), you have many possibilities to choose from. It could be an isosceles triangle with each half-triangle having side lengths [itex]2,3,\sqrt{13}[/itex] by using pythagoras' theorem, which would then be easy to construct.

 

[AlephZero]

You can construct it geometrically if you know this theorem: http://www.mathopenref.com/arccentralangletheorem.html

Draw side a, find the center of the circle and draw it, then draw a line parallel to a at the given altitude.

 

[skrat]

Ugh... altitude = 3cm... I took that as being the side length of the triangle. Sorry, it's been a long night. That's another thing that'll waste everyone's time

Yes but it is a waste of time in a good way. :D

Just once more... Altitude to side a means that the altitude of 3cm connects point A to side a perpendicularly, correct?

YES.

If that is the case (which at this point I wouldn't bet heavily on), you have many possibilities to choose from. It could be an isosceles triangle with each half-triangle having side lengths [itex]2,3,\sqrt{13}[/itex] by using pythagoras' theorem, which would then be easy to construct.

I was afraid you might say that, because I came to the same conclusion (but only in my head with no proof). It's not that I have to proof anything but it just came to my mind that it might be interesting to see the mathematical proof of more possibilities and with it also exactly what are they.

 

[Mentallic]

I was afraid you might say that, because I came to the same conclusion (but only in my head with no proof). It's not that I have to proof anything but it just came to my mind that it might be interesting to see the mathematical proof of more possibilities and with it also exactly what are they.

Actually we should have thought about it some more, because it doesn't look as though it's going to have many possible answers. If you want to prove it for yourself then denote the angle at A in the left triangle by [itex]\theta[/itex] and then the angle at A in the right triangle will be [itex]60^o-\theta[/itex]. Using simple trigonometry you can then find the length of side a on only the left triangle (you can denote it by a₁) and similarly, on only the right triangle (denoted [itex]a_2=4-a_1[/itex]).

You should then have two equations in two unknowns. See if you can find if these two equations intersect anywhere other than when [itex]\theta = 30^o[/itex].EDIT:
It doesn't work when it's an isosceles triangle. [itex]\tan 30^o \neq 2/3[/itex] which would need to be the case.

So I've failed to answer your question thus far. It's been a great night...

Anyway I'm currently working on the algebra. You should too.

 

[pbuk]

Actually we should have thought about it some more, because it doesn't look as though it's going to have many possible answers. If you want to prove it for yourself then denote the angle at A in the left triangle by [itex]\theta[/itex] and then the angle at A in the right triangle will be [itex]60^o-\theta[/itex]. Using simple trigonometry you can then find the length of side a on only the left triangle (you can denote it by a₁) and similarly, on only the right triangle (denoted [itex]a_2=4-a_1[/itex]).

You should then have two equations in two unknowns.

This is exactly what I said in post 6.

 

[pbuk]

You can construct it geometrically if you know this theorem: http://www.mathopenref.com/arccentralangletheorem.html

Draw side a, find the center of the circle and draw it, then draw a line parallel to a at the given altitude.

Brilliant - did you get this by guessing that the angle of 60° was chosen because 60/2 = 30 and 60 + 30 = 90?

 

[Mentallic]

This is exactly what I said in post 6.

I didn't see that at first. You must have edited your post, right? Ugh this is too embarrassing, good night guys.

 

[pbuk]

I didn't see that at first. You must have edited your post, right? Ugh this is too embarrassing, good night guys.

You could redeem yourself by explaining why the numerical value of the angle ## \theta ## is a rational multiple of ## \pi ## degrees.

 

[ehild]

See picture. Draw side a, and a parallel line at distance of 3 cm. The point A should be on that parallel line.

Point A is also on a circle from where the side a looks at the viewing angle of α= 60°. (Angles Subtended by Same Arc Theorem). This circle is circumcircle of the triangle, but also circumcircle of other triangles with α=60° and a =4 cm. That circumcircle can be drawn around the equilateral triangle with base a. The cross-section of the circle and the parallel line gives the point A.

ehild

 

[pbuk]

Brilliant - did you get this by guessing that the angle of 60° was chosen because 60/2 = 30 and 60 + 30 = 90?

Oh, it works for any angle

 

[AlephZero]

Brilliant - did you get this by guessing that the angle of 60° was chosen because 60/2 = 30 and 60 + 30 = 90?

No, I "got it" because when I was at school (a long time ago!), 12-year-olds learned geometry

 

[skrat]

I understand everything except: How on Earth do I get that circle? What is the radius? Where is the center?

 

[skrat]

Do I firstly make a Equilateral triangle with ##a=4cm## and than draw a Circumscribed circle. After that I should draw a line perpendicular to any of the sides and is 3 cm away.

Does this bring me to right solution?

 

[ehild]

I think it does if I understand correctly what you wrote.

Draw the side a, a line 4 cm long. The ends are B and C. Draw an equilateral triangle with base of that line segment. Halve the sides, and connect to the opposite vertices. These lines are also heights. They intersect in the centre of the circumscribed circle. Draw the circle. Measure 3 cm up from the base BC on the height belonging to side BC. Draw a line through that point parallel with BC. The line intersects two points from the circle: these are the third points of the triangle. You get two symmetric triangles.

ehild