Is x + y > z?

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Tue Feb 10, 2009 2:11 am

In a triangle with angles of degree measures x, y, and z, which are opposite sides A, B, and C, respectively, is x + y > z?

(1) x = y

(2) C^2 <A^2 + B^2

IMO B and OA

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
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Source: Beat The GMAT — Data Sufficiency | Tue Feb 10, 2009 2:11 am

billzhao: Senior | Next Rank: 100 Posts; Posts: 80; Joined: Mon Feb 02, 2009 6:36 am; Thanked: 10 times

Re: Is x + y > z?

by billzhao » Wed Feb 11, 2009 2:06 am

sanju09 wrote:In a triangle with angles of degree measures x, y, and z, which are opposite sides A, B, and C, respectively, is x + y > z?

(1) x = y

(2) C^2 <A^2 + B^2

IMO B and OA :

From (1), we cannot tell whether x+y>z
eg. x=10, y=10, z=160, which satisfies: x=y, but x+y<z
x=80, y=80, z=20, which satisfies: x=y, but x+y>z

from (2), we have A^2+B^2>C^..............(m), and according to law of cosines: C^2=A^2+B^2-2*a*b*cos(z).................(n), from (m) and (n), we have A^2+B^2>A^2+B^2-2*a*b*cos(z) => 2*a*b*cos(z) >0. Since a>0 and b>0, we have cos(z)>0, which is z<90 degree. Since x+y+z=180 and z<90, x+y>90>z. So (2) is sufficient.

My answner is (B)

Yiliang

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

Re: Is x + y > z?

by sanju09 » Wed Feb 11, 2009 4:14 am

billzhao wrote:
sanju09 wrote:In a triangle with angles of degree measures x, y, and z, which are opposite sides A, B, and C, respectively, is x + y > z?

(1) x = y

(2) C^2 <A^2 + B^2

IMO B and OA :
From (1), we cannot tell whether x+y>z
eg. x=10, y=10, z=160, which satisfies: x=y, but x+y<z
x=80, y=80, z=20, which satisfies: x=y, but x+y>z

from (2), we have A^2+B^2>C^..............(m), and according to law of cosines: C^2=A^2+B^2-2*a*b*cos(z).................(n), from (m) and (n), we have A^2+B^2>A^2+B^2-2*a*b*cos(z) => 2*a*b*cos(z) >0. Since a>0 and b>0, we have cos(z)>0, which is z<90 degree. Since x+y+z=180 and z<90, x+y>90>z. So (2) is sufficient.

My answner is (B)

Good! But use of Cosine Law could be your personal approach, can you come up with some "Globally Accepted" way to crack it?

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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moorthy76: Senior | Next Rank: 100 Posts; Posts: 53; Joined: Wed Jan 14, 2009 7:58 pm; Location: india; Thanked: 2 times; GMAT Score:700

by moorthy76 » Wed Feb 11, 2009 4:31 am

IMO B...

I applied pythagorus theorem...

Best Regards,
Surya

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Wed Feb 11, 2009 4:46 am

moorthy76 wrote:IMO B...

I applied pythagorus theorem...

Howdy! But where and how did you apply that?

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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moorthy76: Senior | Next Rank: 100 Posts; Posts: 53; Joined: Wed Jan 14, 2009 7:58 pm; Location: india; Thanked: 2 times; GMAT Score:700

by moorthy76 » Wed Feb 11, 2009 5:04 am

I thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. ...

Best Regards,
Surya

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Wed Feb 11, 2009 5:11 am

moorthy76 wrote:I thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. ...

It is still an incomplete elucidation!

You thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. .. which angle? or then what? :!:

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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x2suresh: Master | Next Rank: 500 Posts; Posts: 258; Joined: Thu Aug 07, 2008 5:32 am; Thanked: 16 times

by x2suresh » Wed Feb 11, 2009 7:45 am

sanju09 wrote:
moorthy76 wrote:I thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. ...
It is still an incomplete elucidation!

You thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. .. which angle? or then what? :!:

Here is my theory

IF C^2= A^2+B^2 is true then z=90

Assume A and B lengths are fixed.

If you decrease the angle z.. then opposite side lenght (C) will also decrease.

If you increase the angle z.. then opposite side lenght (C) will also increase.

So C^2< A^2+B^2 --> means z<90.

x+y+z=180 and z<90 --> x+y>90>z

B sufficient.

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