Is the triangle ABC with sides a, b,

Is the triangle ABC with sides a, b, c an acute angles triangle?
1) The Sides AB (length = c) and BC (length = a) are 13 and 17 respectively and the angle ABC is 30 deg.
2) The triangle with sides a^2, b^2 and c^2 is a right-angled triangle.

1) using law of cosine,
c^2= a^2+b^2-2abcos(c)
we can find length and check if it is acute or not
sufficient

2) a,b,c is right angled
sufficient

Ans D

Statement 1:
The two sides (AB and BC) and the included angle (Angle B) are given. We can deduce that the other two angles are fixed as the two sides are also fixed.
Sufficient.

Statement 2:
If the three sides A^2, B^2, C^2 together satisfy the pythagoras theorem then the triangle is a right angled triangle.
Hence, A^4 + B^4 = C^4

Let's assume
A^2 = 3
B^2 = 4
C^2 = 5
So,
3^2 + 4^2 = 5^2

Now,
A = \(\sqrt{3}\)
B =\(\sqrt{4}\)
C =\( \sqrt{5}\)

For a triangle to be acute, A^2 + B^2 > C^2
Here, 3 + 4 > 7
Thus, the given triangle is acute angled triangle.
Hence Sufficient.

Option D

Option B

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient

IMO D

Is the triangle ABC with sides a, b, c an acute angles triangle?

1) The Sides AB (length = c) and BC (length = a) are 13 and 17 respectively and the angle ABC is 30 deg.

Lets draw the altitude AD from A

In triangle ABD, AB=13, Angle ABD=30
So, AD= 13 sin30= 6.5
BD=13cos30= 11.25

AlSo, in triangle ACD, tan Angle DAC= (13-11.25)/6.5=0.88

Therefore , 30< DAC < 45
So angle BAC= 60+DAC = > 90
Sufficient.

2) The triangle with sides a^2, b^2 and c^2 is a right-angled triangle.

Let a^2=3, b^2=4, C^2=5 Forms a right angle triangle

Now , From cosine rule, Cos angle ABC= [(c^2+a^2)-b^2]/2ac = 4/2sqrt15= 2/4=1/2 ~ 60 degree
Similarly, other angles can be found.

Sufficient.

answer
State 1:sufficient
given AC=b,AB=c=13 and BC=a=17,Angle ABC=30
triangle ABC WILL be acute if BC^2<AB^2+AC^2
Now applying cos30=a^2+c^2-b^2/2ac
putting value and solving, we get b^2=458-221sqrt3= approx. 76
conditionBC^2>AB^2+AC^2 is satisfied as 289>13^2+76
State 2:sufficient
from given condition ,let c^2 be the largest side,then c^4=a^4+b^4=(a^2+b^2)^2-2 x a^2 x b^2
c^2<a^2+b^2

IMO E

First statement does not give us enough information about the angles of the triangle, there is a possibility that one of the anglkes can be more than 90 degree.
From the second statement alone we cant get the answer, knowing that a^2, b^2,c^2 makes a right angle triangle does not give us enough information if the triangle with a,b,c is a right angle triangle or not.

Using both statements we cannot conclude anything here.

Is the triangle ABC with sides a, b, c an acute angles triangle?
1) The Sides AB (length = c) and BC (length = a) are 13 and 17 respectively and the angle ABC is 30 deg.
2) The triangle with sides a^2, b^2 and c^2 is a right-angled triangle.

Sum of angles of a triangle = 180

S1: Enough to know that sum of remaining angles will be 150. Could be 90+70 (not acute) or 75+75 (acute)
IS

S2: Square of the sides makes it Pythagoras and since one anglie is 90 deg the other two will add up to 90 deg i.e. their values will also be less than 90 deg.
Suffiecient

Answer is B.

IMO D

For the solution, PFA

1- from statement side a is 4<a<30. Since angle C has to be less than 30 degree , been shorter than side b of angel B equals to 30 degree. Hence taking the max value of angle C +angle b = 29.99+30= 59.99. Angle A = 180-59.99= 120.01.
So not an acute angle triangle

2- Since Pythagoras triangle property is applicable, not an acute angle triangle .

Ans is D

only option 2 is sufficient

Awesome question
Acute triangle ?
1. Two sides and one included angle is given. Follow below method



For this type of triangle, we must use The Law of Cosines first to calculate the third side of the triangle; then we can use The Law of Sines to find one of the other two angles, and finally use Angles of a Triangle to find the last angle.

Law of cosines:

The Law of Cosines (or Cosine Rule) says:
c2 = a2 + b2 − 2ab cos(C)

Law of Sines: (refer above picture)
\(\frac{(a}{sin A)} = \frac{b}{sin B} = \frac{c}{sin C}\)

Angles of a Triangle:

A + B + C = 180°

So, all in all statement 1 is sufficient to know the type of triangle

2. This means that \(a^2 b^2 and c^2\) form a Pythagorean triplet(there are many of them- (3, 4, 5) (5, 12, 13) (7, 24, 25) (8, 15, 17) (9, 40, 41) (11, 60, 61) (12, 35, 37) (13, 84, 85) (16, 63, 65) (20, 21,29) (28, 45, 53) (33, 56, 65) (36, 77, 85) (39, 80, 89) (48, 55, 73) (65, 72, 97)

Square root of these triplet form the sides of the triangle. e.g. triplet - 3,4,5
corresponding sides-\( \sqrt{3}, 2 and \sqrt{5}\)

Now a triangle can be formed using - https://www.mathopenref.com/consttrianglesss.html

But point worth noting here is that - side is \(c ^2 = a^2 + b^2\) that means the angle between sides sides a and b is 90 degrees, but
This does not make us sure that the other angles are acute- refer picture below. Even if other angles are acute or obtuse atleast one angle is 90 degree(triangle is not acute)

Hence, Statement 2 is sufficient
Answer Option D

Is the triangle ABC with sides a, b, c an acute angles triangle?
1) The Sides AB (length = c) and BC (length = a) are 13 and 17 respectively and the angle ABC is 73.8 deg.
2) The triangle with sides a^2, b^2 and c^2 is a right-angled triangle.

an acute ∆ ; a^2+b^2>c^2 is a<=b<=c
all angles are <90*
#1
The Sides AB (length = c) and BC (length = a) are 13 and 17 respectively and the angle ABC is 73.8 deg.
third side angle is 73.8 ; and sum of angle for length c & a ; 106.2 whose sum of can 90+16.2 or it can be 73 +33.2 we get yes and no ; insufficient
#2
The triangle with sides a^2, b^2 and c^2 is a right-angled triangle.
a right angled ∆ a^2+b^2=c^2
such that one of the angles is 90* ; so we can say that a, b, c is not an acute angles triangle.
OPTION B

I don't think trigonometry is part of the GMAT? Is there something here we could learn regardless?

The solution posted here (https://gmatclub.com/forum/is-the-triangle-abc-with-sides-a-b-329024.html#p2566960) is not based on Trignometry.

The question is very tricky and full of traps. Must take into account a strategy

The key for solving a lot of high end Data Sufficiency geometry questions is to determine and know when a figure is fixed and can not move (I believe Manhattan Prep refers to the concept as Rubber Band Geometry).

Generally, if we know definitive values for the dimensions and angles that make up the Triangle Congruency Rules (S-A-S , A-S-A, S-S-S, R-H-S), we can say that the triangle given if FIXED and will not move.

When the triangle is fixed and we have known values that fix the triangle, theoretically anything about that triangle can be determined, even if we can’t do the operations ourselves.


(1) side AB is adjacent to Side BC —-> we are given both lengths (13 and 17)

The Included Angle in between the 2 side lengths is Angle ABC.

Therefore, we are given a fixed value for:

Side of a Triangle (S)

The Included Angle in between adjacent Sides of the Triangle (A)

And the Adjacent Side of the Triangle (S)

Given these fixed values, it would be theoretically possible to determine any other dimension:

We could use the cosine file to find the 3rd side. Then we could use the sine rule proportions to find the other angles.

Since the triangle is fixed in place, the answer will be either a definitive Yes or definitive No.

S1 is Sufficient

(2) again, we are given that the side lengths squared make up a right triangle.

Regardless of the values you put in for the a , b , and c you will always find a triangle in which the inequality

(Longest side)^2 < (shorter side)^2 + (other shorter side)^2

And the triangle will be Acute

D




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