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

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

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New post 06 Feb 2019, 20:00
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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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Re: In an isosceles triangle ABC, if C = 40 degrees, and AB^2 + AC^2 < BC^  [#permalink]

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New post 06 Feb 2019, 20:19
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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

Image



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

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New post 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.

Answer: 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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