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I is a recycled 30:60:90 triangle---> so no obtuse triangle while for II & III it is possible to construct a triangle with atleast one obtuse angle

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
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Re: Which of the following gives a complete set of the triangles [#permalink]

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02 Jan 2013, 13:00

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..

Here's an even more challenging problem along the same lines.

Attachment:

triangle in circle.JPG [ 19.46 KiB | Viewed 2615 times ]

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.

Re: Which of the following gives a complete set of the triangles [#permalink]

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26 Mar 2013, 16:28

1

This post received KUDOS

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My method to determine such problems is by using the pythogoras theorem.

If (largest side)^2 > Sum of square of other 2 sides then the triangle is OBTUSE TRIANGLE If (largest side)^2 = Sum of square of other 2 sides then the triangle is RIGHT ANGLE TRAINGLE If (largest side)^2 < Sum of square of other 2 sides then the triangle is ACUTE TRIANGLE

Tr1 has sides 6-9-10: 100 < 81+36 hence acute Tr. Tr2 has sides 8-14-17: 289 > 196+64 = 260 hence OBTUSE Tr Tr 3 has sides 5-12-14: 196 > 25+144 = 169 hence again OBTUSE Tr.

Re: Which of the following gives a complete set of the triangles [#permalink]

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27 Mar 2013, 21:40

[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.
_________________

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
_________________

Re: Which of the following gives a complete set of the triangles [#permalink]

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28 Mar 2013, 10:29

mikemcgarry wrote:

Here's an even more challenging problem along the same lines.

Attachment:

triangle in circle.JPG

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

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.

Re: Which of the following gives a complete set of the triangles [#permalink]

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28 Mar 2013, 10:30

1

This post received KUDOS

mikemcgarry wrote:

vinaymimani wrote:

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

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.

Hi Mike could you please provide the official answer and explanation to the problem with the circle. Thanks, Raj

Dear Raj

As I discuss in this post: http://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: http://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).

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