Lolaergasheva wrote:

fluke wrote:

PQ=x

QR=x+2

PR=y

Question: is y>x+2

(1)

y=x+3

x+3>x+2

The angle opposite to the longest side of the triangle is greatest.

\(\angle PQR\) is the greatest.

Sufficient.

(2)

x=2;

x+2=4;

2<y<6

If y is about 2; the angle opposite y will be smaller than angle opposite x+2.

If y is about 6; the angle opposite y will be the greatest.

Not Sufficient.

Ans: "A"

You are great as always. But I didn't understand why you suppose y>x+2 ?

No, I didn't suppose so. I just rephrased what's asked in the question. The question asks

which angle is the greatest among the three angles.

What sides are there;

x,x+2,y

What angles are opposite these sides;

opposite x -> \(\angle QRP\)

opposite x+2 -> \(\angle QPR\)

opposite y -> \(\angle PQR\)

x+2 will always be greater than x. So; angle opposite to side x+2 will be greater than the

angle oppsite to x.

\(\angle QPR\) will always be greater than \(\angle QRP\). We need to know about

\(\angle PQR\).

The only question then stands, "Is y> x+2 or y<x+2"

If y> x+2; then y becomes the longest side and the angle opposite to it will become the greatest angle.

If y< x+2; then x+2 becomes the longest side and the angle opposite to it will become the greatest angle.

That's all!!

1) This statements tells us that y=x+3

For any value of x;

y will be the longest side and the angle oppsite y will be the greatest.

2) This statement tells us that;

x=2

x+2=4

y is any value between 2 and 6(if one side of the triangle is 2 and other side 4; the third side will be between the difference

and sum of the other two sides)

means

(4-2) < y < (4+2)

2<y<6

y can be 3 which is less than x+2. x+2 becomes longest.

y can be 5 which is greater than x+2. y becomes longest.

So; we don't definitely know whether y>x+2.

Not sufficient.

_________________

~fluke

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