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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
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c^2 is greater than a^2+b^2 then the angel opposite c must be greater than 90


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.
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
Bunuel wrote:
I attached the diagram, which was missing in initial post.

Attachment:
Trianlge.PNG
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=4^2, since a^2+b^2<15<16 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.

Hope it helps.


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?
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
fozzzy wrote:
Bunuel wrote:
I attached the diagram, which was missing in initial post.

Attachment:
Trianlge.PNG
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=4^2, since a^2+b^2<15<16 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.

Hope it helps.


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?


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)
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
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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!
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
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smily_buddy wrote:
Attachment:
Trianlge.PNG
In the triangle above, is x > 90?

(1) a^2 + b^2 <15
(2) c > 4



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.
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
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
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
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longhaul123 wrote:
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


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.
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In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
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smily_buddy wrote:

In the triangle above, is \(x > 90\)?

(1)\( a^2 + b^2 <15\)
(2) c > 4

Attachment:
triangle.GIF


(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.
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
Very good question! Thanks for posting.
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]
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