# Thread: Equiangular Triangle w/ 30/60/90 degree angle

1. ## Equiangular Triangle w/ 30/60/90 degree angle

Hey Guys,

So I have this Equiangular Triangle, NR = 6 and NR bisects angle MNQ and QR Bisects angle MQN

Since this an Equiangular Triangle. All of its sides are 60 degrees.
When both these sides are bisected, they turn into 30 degree angles.

After this step, I'm lost. I understand that this Triangle has to be a 30-60-90 degree angle since that's what this section is about. Angle R would most likely be my 90 degree angle. But which angle is my 60 degree angle? Since both angles are bisected, wouldn't both of them be 30 degrees?

The answer on the back of the book is 6 sqrt 3.

2. ## Re: Equiangular Triangle w/ 30/60/90 degree angle

The smaller triangle is isosceles, with the two equal angles 30 degrees, which means the last angle must be 120 degrees (why?)

Now that you have two sides and the angle between them, you can use the cosine rule to find NQ.

3. ## Re: Equiangular Triangle w/ 30/60/90 degree angle

Hi,
Here's a purely geometric derivation -- no trig required.

4. ## Re: Equiangular Triangle w/ 30/60/90 degree angle

Originally Posted by johng
Hi,
Here's a purely geometric derivation -- no trig required.

Those results require trigonometry.

5. ## Re: Equiangular Triangle w/ 30/60/90 degree angle

hi proveit,
Exactly what portions of my previous post require trigonometry? The only thing I can think of is that you believe the property of a 30-60-90 right triangle that was used requires trig. I refute this; c.f. Special right triangles - Wikipedia, the free encyclopedia

6. ## Re: Equiangular Triangle w/ 30/60/90 degree angle

Originally Posted by johng
hi proveit,
Exactly what portions of my previous post require trigonometry? The only thing I can think of is that you believe the property of a 30-60-90 right triangle that was used requires trig. I refute this; c.f. Special right triangles - Wikipedia, the free encyclopedia
Thanks guys.
Yeah, we haven't covered Trigonometry in this section yet. I was thinking of making a perpendicular line from R to NQ but I wasn't so sure if I could do that for this problem.