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# In an isosceles triangle ABC, if C = 40 degrees, and AB^2 + AC^2 < BC^

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e-GMAT Representative
Joined: 04 Jan 2015
Posts: 3074
In an isosceles triangle ABC, if C = 40 degrees, and AB^2 + AC^2 < BC^  [#permalink]

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06 Feb 2019, 20:00
00:00

Difficulty:

15% (low)

Question Stats:

81% (01:31) correct 19% (01:38) wrong based on 52 sessions

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In an isosceles triangle ABC, if ∠C = 40 degrees, and $$AB^2 + AC^2 < BC^2$$, then what is the difference between the largest and the smallest angles of the triangle?

A. 30
B. 40
C. 60
D. 70
E. 100

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Joined: 02 Aug 2009
Posts: 7960
Re: In an isosceles triangle ABC, if C = 40 degrees, and AB^2 + AC^2 < BC^  [#permalink]

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06 Feb 2019, 20:19
1
1
EgmatQuantExpert wrote:
In an isosceles triangle ABC, if ∠C = 40 degrees, and $$AB^2 + AC^2 < BC^2$$, then what is the difference between the largest and the smallest angles of the triangle?

A. 30
B. 40
C. 60
D. 70
E. 100

Good question...

Few rules..

Let sides be a, b and c
(I) At $$a^2+b^2=c^2$$, the triangle is right angled triangle
(II) If $$a^2+b^2>c^2$$, the triangle is acute angled triangle, that is all angles are acute ( less than 90)
(III) If $$a^2+b^2<c^2$$, the triangle is obtuse angled triangle, that is one angle is greater than 90.

Here we are in (III) case, so we should surely have an angle greater than 90.

Now with this info, let us check the possible angles of the triangle
(A) If 40 is the angle made twice ---- 40, 40, 100
(B) If 40 is the angle made only once ---- 40, $$\frac{180-40}{2}$$, $$\frac{180-40}{2}$$-----40,70,70

In case (I) we have an angle > 90.. hence angles are 40,40,100

We are looking for the difference between the largest and the smallest angles of the triangle=100-40=60

C
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Joined: 04 Jan 2015
Posts: 3074
Re: In an isosceles triangle ABC, if C = 40 degrees, and AB^2 + AC^2 < BC^  [#permalink]

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08 Feb 2019, 00:10

Solution

Given:
• ABC is an isosceles triangle
• ∠C = 40 degrees
• $$AB^2 + AC^2 < BC^2$$

To find:
• The difference between the largest and smallest angles of triangle ABC

Approach and Working:
We are given that, $$AB^2 + AC^2 < BC^2$$, which implies ABC is an obtuse angles triangle at A.
• Thus, ∠A > 90 degrees

So, for ABC to be an isosceles triangle, the only possibility is that ∠C = ∠B = 40 degrees
• Thus, ∠A = 180 – 40 – 40 = 100 degrees

Therefore, the difference between the largest and smallest angles of triangle ABC, ∠A - ∠B = 100 – 40 = 60 degrees

Hence the correct answer is Option C.

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Re: In an isosceles triangle ABC, if C = 40 degrees, and AB^2 + AC^2 < BC^   [#permalink] 08 Feb 2019, 00:10
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