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Is Right angle triangle?

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by manpsingh87 » Tue May 31, 2011 10:00 am
hardikm wrote:Is triangle ABC a right triangle?
1. < ABC = < BCA
2. <ABC = <CAB + 30 degrees

Source: GMAT Club App
<ABC=x;
<BCA=y;
<CAB=z;
1) x=y;
also we know that x+y+z=180;
2x+z=180;
z=180-2x; when x=45; z=90;
and when x=30; z=120;
hence 1 alone is not sufficient to answer the question.

2)x=z+30;
x+y+z=180;
z+30+y+z=180;
2z+y=180-30;
y=150-2z;
when z=30; y=90;
when z=45; y=150-90=60; x=45+30=75;
hence 2 alone is not sufficient to answer the question.

combining 1 and 2 we have;
2x+z=180;
also x=z+30;
2(z+30)+z=180;
3z+60=180;
3z=120;
z=40;
x=70=y;
i.e. ABC is not a right angle triangle hence C
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by Stuart@KaplanGMAT » Tue May 31, 2011 10:06 am
hardikm wrote:Is triangle ABC a right triangle?
1. < ABC = < BCA
2. <ABC = <CAB + 30 degrees

Source: GMAT Club App
Let's answer the question using minimal math, relying on our general knowledge of triangle rules.

Rephrasing the question: is one of the angles of triangle ABC 90 degrees?

1) two of the angles are equal. Neither of those angles could be 90 (since we only have a total of 180 degrees), but we know nothing about the third angle: insufficient.

2) one angle is 30 degrees less than another angle. One of those could be 60, one could be 90, or we could pick pretty much any other random angles and not end up with 90 degrees: insufficient.

Together: two are equal and those two angles are each 30 degrees less than the third angle.

Well, a right isosceles triangle is 45/45/90. Since we need a 30 degree difference, we definitely don't have a 45/45/90. Accordingly, our triangle is definitely NOT right: sufficient, choose (C).
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by cans » Wed Jun 01, 2011 4:25 am
a) < ABC = < BCA => isosceles triangle. Insufficient.
b) <ABC = <CAB + 30 degrees insufficient.
a&b) <ABC + <CAB + < BCA = 180
2(<CAB + 30) + <CAB = 180 => <CAB = 40
thus <ABC = <BCA = 70
Not right angled
Sufficient
ImO C
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by Stuart@KaplanGMAT » Wed Jun 01, 2011 10:14 pm
manpsingh87 wrote: combining 1 and 2 we have;
2x+z=180;
also x=z+30;
2(z+30)+z=180;
3z+60=180;
3z=120;
z=40;
x=70=y;
i.e. ABC is not a right angle triangle hence C
I just wanted to address this solution, since it makes a common mistake for which test takers need to be watchful. Remember, when solving an isosceles triangle you can't assume which angles are equal and which one is different.

Another perfectly acceptable solution to the problem is:

x + 2z = 180
x = z + 30
3z + 30 = 180
3z = 150
z = 50
x = 50 + 30
x = 80

So, we could also satisfy all the rules with an 80/50/50 triangle.

On this particular question it doesn't matter (since neither solution has a 90 degree angle), but there's at least one other question that I've seen on which there are two solutions for an isosceles triangle and the less-commonly solved one changes the answer to the question.
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by cans » Thu Jun 02, 2011 8:34 am
Stuart Kovinsky wrote:
Another perfectly acceptable solution to the problem is:

x + 2z = 180
x = z + 30
3z + 30 = 180
3z = 150
z = 50
x = 50 + 30
x = 80

So, we could also satisfy all the rules with an 80/50/50 triangle.

On this particular question it doesn't matter (since neither solution has a 90 degree angle), but there's at least one other question that I've seen on which there are two solutions for an isosceles triangle and the less-commonly solved one changes the answer to the question.
Hi Stuart, This solution is not acceptable as according to Manpsingh87 <ABC=x;
<BCA=y;
<CAB=z;
and as given in question, <ABC=<BCA or x=y and thus it has to be 2x+z=180
Is triangle ABC a right triangle?
1. < ABC = < BCA
2. <ABC = <CAB + 30 degrees
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by Stuart@KaplanGMAT » Thu Jun 02, 2011 11:57 am
You're 100% right! That will teach me not to read all the information carefully!
cans wrote:
Stuart Kovinsky wrote:
Another perfectly acceptable solution to the problem is:

x + 2z = 180
x = z + 30
3z + 30 = 180
3z = 150
z = 50
x = 50 + 30
x = 80

So, we could also satisfy all the rules with an 80/50/50 triangle.

On this particular question it doesn't matter (since neither solution has a 90 degree angle), but there's at least one other question that I've seen on which there are two solutions for an isosceles triangle and the less-commonly solved one changes the answer to the question.
Hi Stuart, This solution is not acceptable as according to Manpsingh87 <ABC=x;
<BCA=y;
<CAB=z;
and as given in question, <ABC=<BCA or x=y and thus it has to be 2x+z=180
Is triangle ABC a right triangle?
1. < ABC = < BCA
2. <ABC = <CAB + 30 degrees
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