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 :
Is x + y > z?
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- sanju09
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From (1), we cannot tell whether x+y>zsanju09 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 :
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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Good! But use of Cosine Law could be your personal approach, can you come up with some "Globally Accepted" way to crack it?billzhao wrote:From (1), we cannot tell whether x+y>zsanju09 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 :
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)
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
Sanjeev K Saxena
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- sanju09
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Howdy! But where and how did you apply that?moorthy76 wrote:IMO B...
I applied pythagorus theorem...
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
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Sanjeev K Saxena
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It is still an incomplete elucidation!moorthy76 wrote:I thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. ...
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
Sanjeev K Saxena
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sanju09 wrote:It is still an incomplete elucidation!moorthy76 wrote:I thought side c^2 <A^2 +B^2, then the angle is not 90 degrees. ...
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.