Guys i don't understand this concept very well, please correct me.

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- Mar 6th 2011, 05:50 AMvaironxxrdSpecial right triangle 60,30,90
Guys i don't understand this concept very well, please correct me.

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

Uploaded with ImageShack.us - Mar 6th 2011, 05:55 AMe^(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 $\displaystyle 2a=5 \to a = 2.5$

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

edit: it's even easier if you're allowed to use trig - Mar 6th 2011, 06:18 AMvaironxxrd
- Mar 6th 2011, 07:04 AMe^(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 $\displaystyle x : x\sqrt{3} : 2x$

Your ratio is $\displaystyle a : b : 5$

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

Once you've found x you can sub it into the first two parts to find*a*and*b* - Mar 6th 2011, 07:35 AMvaironxxrd
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, 12:12 PMbjhopper
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