EgmatQuantExpert
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