triangle construction: 60, 45, 75-deg angles, 7in perim

[soccerisgreat]

I've been asked to construct a triangle with a 60, 45, and 75 degree angles and a perimeter of 7 inches. I can construct the 60 degree angle but don't know where to go from there. HELP!

 

[Deleted member 4993]

Re: triangle construction

I've been asked to construct a triangle with a 60, 45, and 75 degree angles and a perimeter of 7 inches. I can construct the 60 degree angle but don't know where to go from there. HELP!

If I were to do this problem:

I would find the length of the side opposite to 75° - using sine law.

Then construct 60° and 45° angles on that side to get the triangle.

Please show your work/thoughts - tell us exactly where you are stuck - so that we know where to begin to help you.

 

[Denis]

Re: triangle construction

If I were to do this problem:
I would find the length of the side opposite to 75° - using sine law.

Just curious: why not the side opposite 60, since COS(60) = 1/2 ?

Also, since (wonder why!) perimeter is to be 7, you could start
by letting side opposite 60 = 1, calculate the 2 other sides (simple enough),
then bring the whole shebang proportionally up to 7: no?

 

[Deleted member 4993]

Very legitimate and correct yet different ways of attacking the same problem!

The student wanted to construct - I assumed withcompass and ruler. My logic was that it is relatively easier to draw 60° and 45° (may not be in reality - just my perception). So Find the base and draw the base angles - intersection gives you the apex of the triangle.

 

[soroban]

I agree with Denis . . .

                    *
                  * |*
                *   | *
              *     |  * 2a/√3
       a√2  *     a |   *
          *         |    *
        *           |     *
      * 45°         |  60° *
    * * * * * * * * * * * * *
    :       a       : a/√3  :

Construct a 45-45-90 right triangle with sides: \(\displaystyle a,\:a,\:a\sqrt{2}\)

Conjoin a 30-60-90 right triangle with sides: \(\displaystyle \frac{a}{\sqrt{3}},\:a,\:\frac{2a}{\sqrt{3}}\)

The perimeter is: \(\displaystyle \:a\,+\,\frac{a}{\sqrt{3}}\,+\,\frac{2a}{\sqrt{3}}\,+\,a\sqrt{2} \;=\;a\,+\,a\sqrt{3}\,+\,a\sqrt{2}\)

So we have: \(\displaystyle \;a\left(1\,+\,\sqrt{3}\,+\,\sqrt{2}\right) \;=\;7 \;\;\Rightarrow\;\;a \;=\;\frac{7}{1\,+\,\sqrt{3}\,+\,\sqrt{2}}\)

. . which rationalizes to: \(\displaystyle \:a\;=\;\frac{7}{4}\left(2\,+\,\sqrt{2}\,-\,\sqrt{6}\right)\)

 

[soccerisgreat]

The student wanted to construct - I assumed withcompass and ruler

Subhotosh Khan is correct. I need to construct with a straight edge and compass. It seemed easy to start with the 60 degree angle because I know how to construct that. Unfortunately, I do not know how to construct the 45 degree angle? I can do this on Geometer sketchpad if I understand how to construct the two angles. Then sizing it is easy.

 

[soroban]

Hello, soccerisgreat!

I do not know how to construct the 45 degree angle.

Construct a right angle, and bisect it.
. . or ... construct a square, then draw a diagonal.

 

[soccerisgreat]

Thanks soroban! So simple, why didn't I think of that!?