# Special right triangle 60,30,90

• Mar 6th 2011, 06:50 AM
vaironxxrd
Special right triangle 60,30,90
Guys i don't understand this concept very well, please correct me.

Image Down

http://img560.imageshack.us/img560/5...ighttriang.png

• Mar 6th 2011, 06:55 AM
e^(i*pi)
I prefer to think of it as an equilateral triangle which has been bisected - the image above is one of those halves.

Given that an equilateral triangle has angle 60 degrees and (in this case) side 5 then it follows that side a is half the length it was before being bisected, in other words $2a=5 \to a = 2.5$

If $a=2.5$ then b can be found with Pythagoras theorem.

edit: it's even easier if you're allowed to use trig
• Mar 6th 2011, 07:18 AM
vaironxxrd
Quote:

Originally Posted by e^(i*pi)
I prefer to think of it as an equilateral triangle which has been bisected - the image above is one of those halves.

Given that an equilateral triangle has angle 60 degrees and (in this case) side 5 then it follows that side a is half the length it was before being bisected, in other words $2a=5 \to a = 2.5$

If $a=2.5$ then b can be found with Pythagoras theorem.

edit: it's even easier if you're allowed to use trig

Things is that our teacher wants us to use this formula
• Mar 6th 2011, 08:04 AM
e^(i*pi)
Ah, I didn't see the ratio clearly first time around. Your working looks fine, what don't you understand?

The ratio given is $x : x\sqrt{3} : 2x$

Your ratio is $a : b : 5$

The component parts must also be equal: $a=x : b = x \sqrt{3} : 5 = 2x$

Once you've found x you can sub it into the first two parts to find a and b
• Mar 6th 2011, 08:35 AM
vaironxxrd
Quote:

Originally Posted by e^(i*pi)
Ah, I didn't see the ratio clearly first time around. Your working looks fine, what don't you understand?

The ratio given is $x : x\sqrt{3} : 2x$

Your ratio is $a : b : 5$

The component parts must also be equal: $a=x : b = x \sqrt{3} : 5 = 2x$

Once you've found x you can sub it into the first two parts to find a and b

i just don't understand for example how do i know when do i have to actually do Equations and when do i actually just have to replace X for given number.

My teacher explained to me if the hypotenuse is a whole number we gotta solve. But if it is the short leg ( For 60-30-90) Then we just place X and we are done.. Is this right or did i understand wrong?

Thing is he explains different from every explanations i see online.
• Mar 6th 2011, 01:12 PM
bjhopper
Hi,
In my opinion every geometry student should know the properties of a30-60-90 triangle and it should used directly without using the pythagorean theorem

bjh