In the triangle above, is x 90? (1) a^2 + b^2 4

Question

https://gmatclub.com/forum/download/file.php?id=36945 In the triangle above, is x > 90? (1) a^2 + b^2 <15 (2) c > 4 triangle.GIF

Bunuel · Accepted Answer

https://gmatclub.com/forum/download/file.php?id=36945 In the triangle above, is x > 90? (1) a^2 + b^2 <15 (2) c > 4 Each statement alone is clearly insufficient. When taken together: If angle x were 90 degrees than we would have a^2+b^2=c^2 , since a^2+b^2<15<(16=c^2) then angle x must be greater than 90 degrees (c^2 is greater than a^2+b^2 then the angel opposite c must be greater than 90). Answer: C. P.S. If the lengths of the sides of a triangle are a, b, and c, where the largest side is c, then: For a right triangle: a^2 +b^2= c^2 . For an acute (a triangle that has all angles less than 90°) triangle: a^2 +b^2>c^2 . For an obtuse (a triangle that has an angle greater than 90°) triangle: a^2 +b^2

Bunuel · Answer

If c^2 were equal to a^2+b^2 then we would have a^2+b^2=c^2, which would mean that angle x is 90 degrees. Now, since c^2 is more than a^2+b^2, then angle x, which is opposite c, must be more than 90 degrees: try to increase side c and you'll notice that angle x will increase too.

Hope it's clear.

SergeNew · Answer

Hi Guys,

Would anyone be able to explain why the angle is greater than 90 if c^2 is greater than a^2+b^2?

Serge.

fozzzy · Answer

What would happen is the statement was c>3? how would this question be framed such that the angle could be always below 90 degrees. Since in this particular question the solution will always be greater, what would be the opposite case?

Vips0000 · Answer

If you reverse the values, such as:
(1) a^2 + b^2 >16
(2) c < 4
You would get a case where angle x would be less than 90, provided a triangle is still formed. (Note that a+b will tend to be bigger and C tends to be smaller in this option)

mau5 · Answer

Just a handy tool to prove that the answer is C.

c^2 = a^2+b^2-2abCos(c). In this case , Angle c = x.

Now, we know that c^2 >16

Also, a^+b^2<15.

Thus a^2+b^2<c^2 = a^2+b^2-c^2 <0

a^2+b^2-c^2 = 2abCos(x) ; 2abCos(x) <0. As ab!=0,Cos(x)<0. Thus, X>90 degree.

For c=90 degree, the above equality gives the famous theorem!

greatps24 · Answer

If a^2 + b^2 = c^2 then x= 90.
But , as given in st2, minimum value of C = 5, then C^2 = 25.
It means a^2 + b^2 < c^2.Thus, x <90. Therefore, C.

longhaul123 · Answer

The inequality a^2+b^2,15 shows that the angle x is not equal to 90 so it can lesser or greater. So the statement 1 is sufficient. I know that i have some mistake with the concept, can someone please correct me. thank you

Bunuel · Answer

How does a^2 + b^2 < 15 imply that x is not 90 degrees? We are given that the sum of the square of the lengths of two sides is less than some number. We cannot deduce anything from this. If a = b = 1 and  c = \sqrt{2} , then  a^2 + b^2 = c^2 , which will make x equal to 90 degrees but if a = b = c = 1, then x will be equal to 60 degrees.

Basshead · Answer

(1) a = b = c = 1; each side would be 60 degrees. a = b = 1 and c = 2; in this case x > 90. INSUFFICIENT. (2) No information about either a or b; INSUFFICIENT. (1) We know a^2 + b^2 < 15 and c > 4 c^2 > 16 a^2 + b^2 < 15 < c^2 x is greater than 90 degrees. SUFFICIENT. Answer is C.

dsteeps · Answer

Very good question! Thanks for posting.

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In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4

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