Which of the following gives a complete set of the triangles

in case of II & III
sum of square of 2 sides is less than square of 3

Dear mansoorfarooqui,
You answered the question correctly, but with all due respect, a 6-9-10 triangle has absolutely nothing to do with a 30-60-90 triangle. The sides of the latter involve irrational number ratios, no matter what numbers are chosen for the lengths. The 6-9-10 triangle is an triangle with three acute angles.
Mike

hi mikemcgarry,

thanks for the clarification.. 6-9-10 is not a recycled triangle.. 6-8-10 would have been one in which one angle would be 90 and other two have to be acute... so my mistake..

Best Regards,
Mansoor

Here's an even more challenging problem along the same lines.
In the diagram above, AB = 10 is the diameter of the circle, and AC = 6. Given that point C is inside the circle, which could be the length of BC?
I. 7
II. 8
III. 9

(A) I
(B) II
(C) III
(D) I & II
(E) II & III

The information at the blog link above will help to solve this question. I will post an OA if folks are curious.

Mike

[quote="mikemcgarry"]Consider the following three triangles

I. a triangle with sides 6-9-10
II. a triangle with sides 8-14-17
III. a triangle with sides 5-12-14

Which of the following gives a complete set of the triangles that have at least one obtuse angle, that is, an angle greater than 90°?

(A) I
(B) II
(C) III
(D) I & II
(E) II & III

Basically, there is just one formula for questions like these. For any triangle,\(c^2 = a^2 + b^2 - 2abCos(C)\), where a,b,c and angles A,B,C follow the normal convention. Now, for an acute triangle, the value of 0<Cos(C)<1 --> \(a^2+b^2>c^2\). For obtuse angles, -1<Cos(C)<0 -->\(c^2>a^2+b^2\) and for C = 90 degrees, we have Pythagoras theorem. Trigonometry, is not tested on GMAT(Quoting Bunuel on this), yet doesn't hurt to know some thing useful.

Dear vinaymimani, you obviously have a strong background in math and have studied trigonometry already, so for you, thinking about these questions using the Law of Cosines is perfectly fine. I would caution you, though --- some people studying for the GMAT haven't done any math at all since high school Algebra II ---- these folks are struggling just to remember basic algebra & geometry, have never even met Cosine, and have no earthly clue even what trigonometry is. I guess I would caution you against cavalierly saying, "[it] doesn't hurt to know some thing useful" ----- if folks who are struggling to remember even how to solve for x or what properties a triangle has also are led to believe that there's something about the Law of Cosines they need to understand, that very much could hurt both their studying and their fragile mathematical self-confidence. Always remember --- there's a very wide audience for anything you post on this site.
With great respect,
Mike

Answer (A).

The Line BC would have maximum length when the opposite angle CAB would be as big as possible. In order to increase this angle the point C could touch the circle at some point. At that point we will have Right angle triangle with right angle at ACB.

BC = sqrt (100 - 36)
BC = 8
So BC can be less than 8
only 7 satisfies.

Thanks.

You seem to have missed the entire point of my post. I had written "Trigonometry is NOT tested on GMAT". That actually means my post was just an additional way of solving. And as for the wide audience reading this post, I personaly feel there is nothing wrong in sharing something new, as long as it is not WRONG. There is nothing cavalier about this post. Just a matter of choice for the thousands who are on this site.

With utmost respect.

Dear Raj

As I discuss in this post:
https://magoosh.com/gmat/2012/re-thinkin ... le-obtuse/
we can use the Pythagorean Theorem to determine, from the three sides of a triangle, whether the triangle is right, acute, or obtuse. I'll ask you to look at that page for the detailed discussion, but the nutshell summary is:
If a^2+ b^2 = c^2, then the triangle is a right triangle.
If a^2+ b^2 > c^2, then the triangle is a acute triangle.
If a^2+ b^2 < c^2, then the triangle is a obtuse triangle.
In all of these, of course, I am assuming a < b < c.

We have to combine this with another geometry idea ----- if the vertex an angle is on a circle, and the rays of angle pass through the two endpoints of a diameter of the circle, then the angle is a right angle. See this blog:
https://magoosh.com/gmat/2012/inscribed- ... -the-gmat/

If BC = 8, then 6^2 + 8^2 = 10^2, and it's a right angle, which would mean point C would be on the circle, instead of inside the circle. If we made BC bigger than 8, that would push C further away, outside of the circle. Making BC smaller than 8 pulls it into the circle, so 7 is the only possible length for BC, and the OA for this question is (A).

Mike

While the traditional Pythagorean Theorem applies only to right triangles, this "expanded Pythagorean Theorem" would help us figure out whether any triangle is acute, right, or obtuse. Here are a couple more posts about fact relevant, not just to special triangle, but to all triangles.
https://magoosh.com/gmat/2012/facts-abou ... -the-gmat/
https://magoosh.com/gmat/2013/gmat-math- ... -triangle/
Mike

Solution:

Recall that if a, b, and c are the side lengths of a triangle where c is the longest side length and:

1) a^2 + b^2 = c^2, then the triangle is a right triangle.

2) a^2 + b^2 > c^2, then the triangle is an acute triangle.

3) a^2 + b^2 < c^2, then the triangle is an obtuse triangle.

Since only obtuse triangles could have an obtuse angle, we are determining whether the given triangles are obtuse or not. Of course, if they are, case 3 will be satisfied. Otherwise, it will not.

I. a triangle with sides 6-9-10

We see that 10 is the longest side and we need to determine whether 6^2 + 9^2 < 10^2.

36 + 81 < 100 ?

117 < 100 ?

We see that 117 is NOT less than 100, so the triangle with sides 6-9-10 can’t be an obtuse triangle and thus it won’t contain an obtuse angle.

II. a triangle with sides 8-14-17

We see that 17 is the longest side, and we need to determine whether 8^2 + 14^2 < 17^2.

64 + 196 < 289 ?

260 < 289 ?

We see that 260 is INDEED less than 289, so the triangle with sides 8-14-17 is an obtuse triangle, and thus it will contain an obtuse angle.

III. a triangle with sides 5-12-14

We see that 14 is the longest side, and we need to determine whether 5^2 + 12^2 < 14^2.

25 + 144 < 196 ?

169 < 196 ?

We see that 169 is INDEED less than 196, so the triangle with sides 5-12-14 is an obtuse triangle, and thus it will contain an obtuse angle.

Answer: E

a) Right triangles: c^2 = a^2 + b^2
b) Acute triangles: a^2 + b^2 > c^2
c) Obtuse triangles: a^2 + b^2 < c^2

i) 36 + 81 > 100 NO
ii) 64 + 196 < 289 YES
iii) 25 + 144 < 196 YES

E is the answer.

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