Is x + y > z?

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)

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?

IMO B...

I applied pythagorus theorem...

moorthy76 wrote:IMO B...

I applied pythagorus theorem...

Howdy! But where and how did you apply that?

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

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? :!:

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.