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

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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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17 Aug 2007, 13:09
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Trianlge.PNG [ 1.74 KiB | Viewed 8677 times ]
In the triangle above, is x > 90?

(1) a^2 + b^2 <15
(2) c > 4
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Feb 2012, 03:57, edited 2 times in total.
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17 Aug 2007, 13:25
smily_buddy wrote:
ABC with a trainagle with three sides abc and one of the angle is x

In the triangle above, is x > 90?
(1) a2 + b2 <15
(2) c> 4
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

E for me.
(1) don't know c
INSUFFICIENT

(2) Only know c
INSUFFICIENT

Together, if a^2 + b^2 < 15
We know that at least one angle in the triangle is greater than 90 if a=2.99, b=3.99, c=5. But other angles are less than 90. We don't know where x is; thus, insufficient.
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17 Aug 2007, 14:34
smily_buddy wrote:
ABC with a trainagle with three sides abc and one of the angle is x

In the triangle above, is x > 90?
(1) a2 + b2 <15> 4
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

Note that for a right triangle, c^2 = a^2+b^2, where c is the length of the side opposite the right angle C.

If angle C were less than 90º, c^2<a>a^2+b^2

Together, we are given that b^2 +a^2< c^2
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17 Aug 2007, 17:31
kevincan wrote:
smily_buddy wrote:
ABC with a trainagle with three sides abc and one of the angle is x

In the triangle above, is x > 90?
(1) a2 + b2 <15> 4

Note that for a right triangle, c^2 = a^2+b^2, where c is the length of the side opposite the right angle C.

If angle C were less than 90º, c^2<a>a^2+b^2

Together, we are given that b^2 +a^2< c^2

thats right but we do not know which angle is x? correct?

it could be the largest one or any one. if it is the largest one then x > 90 and vice versa.

so I go with E.
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18 Aug 2007, 00:46
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We don't have to know which angle is x.
Question is asking, whether any of the angle in the triangle is greater than 90.

since c^2 > a^2 + b^2 , we know tht angle opposite to side c will be greater than 90.

So, both the statements are required to answer the question.
My choice is C.
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18 Aug 2007, 00:54
fatal1ty wrote:
We don't have to know which angle is x.
Question is asking, whether any of the angle in the triangle is greater than 90.

since c^2 > a^2 + b^2 , we know tht angle opposite to side c will be greater than 90.

So, both the statements are required to answer the question.
My choice is C.

Together, we are told that the triangle is obtuse- one interior angle is greater than 90º. However, we don't know which which angle is being asked about
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18 Aug 2007, 10:48
fatal1ty wrote:
We don't have to know which angle is x.
Question is asking, whether any of the angle in the triangle is greater than 90.

since c^2 > a^2 + b^2 , we know tht angle opposite to side c will be greater than 90.

So, both the statements are required to answer the question.
My choice is C.

Its not asking whether any of the angle in the triangle is greater than 90.
It is asking that one of the angle is x, Is x > 90? Means one perticular angle.N we dont know which is the one ?

Both Insufficient !

I
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Re: ABC with a trainagle with three sides abc and one of the [#permalink]

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09 Feb 2012, 00:27
According to my doc, OA is C.
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Re: ABC with a trainagle with three sides abc and one of the [#permalink]

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09 Feb 2012, 01:13
E is the correct answer. Both statements are insufficient.
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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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09 Feb 2012, 03:57
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Expert's post
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I attached the diagram, which was missing in initial post.

Attachment:

Trianlge.PNG [ 1.74 KiB | Viewed 8651 times ]
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).

Hope it helps.
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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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17 Feb 2012, 00:26
Quote:
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]

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17 Feb 2012, 02:27
Whenever Asqr + Bsqr Is equal to Csqr than angle X is 90 degrees
Look at both options carefully ,

Asqr + Bsqr Is less than 15 and C greater than 4 means any value more than 4 for c and will automatically increase angle X bcoz as C increases angle 8 increases

Poor in making posts

Hope it will help.

Posted from my mobile device
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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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17 Feb 2012, 02:42
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Expert's post
SergeNew wrote:
Quote:
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.

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.
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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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26 Jan 2013, 21:15
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).

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]

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26 Jan 2013, 21:50
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).

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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28 Jan 2013, 23:31
Expert's post
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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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24 Feb 2013, 21:46
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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: ABC with a trainagle with three sides abc and one of the [#permalink]

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24 Feb 2013, 23:14
abhi47 wrote:
E is the correct answer. Both statements are insufficient.

from one, possible values of (a,b) = (1,2) or (1,3) or (2,3) or (2,2)
from 2nd, given that c>4
Property of triangle is = sum of two sides > third side
therefor only possible values are a= 2, b=3 and c = 5 or a = 3 and b = 2 and c =5
since a^2 + b ^2 = c^2
hence it is a right angle triangle
hence both conditions required to answer
so option(c)
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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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30 Sep 2014, 00:29
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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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01 Oct 2015, 17:41
a^2+b^2 = c^2 then angle opposite of c is right angle triangle
a^2+b^2 < c^2 then angle opposite of c is greater than 90
a^2+b^2 > c^2 then angle opposite of c is less than 90

Using the above logic, we can make of use of both the statements to answer the question.
Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4   [#permalink] 01 Oct 2015, 17:41
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