This is a very deceptive question, because you can't actually assume that the vertical line is a perpendicular bisector; there isn't a right angle in the figure that indicates that \(a\) is 90 degrees, though it certainly can be.

You can drag that vertical line along the horizontal line \(XZ\), making \(a\) obstuse, right, or acute. But in all three scenarios, \(a > c\).

In short, \(0 < c < 60\), since \(XYZ\) is an equilateral with all interior angles equal to 60. Since the 60 degree angle is being divided into two different angles, \(c\) and \(60-c\), \(c\) cannot be 0 or 60, otherwise the vertical line wouldn't exist.

On the other hand, \( 60 < a < 120 \), because any triangle you try to form with \(a\) as an interior angle will include 60 and \(60-c\).

With these two inequalities:

\(0 < c < 60\)
\( 60 < a < 120 \)

We can see that Quantity A is greater than B.


But in fact, we can prove it rigorously as well.

We know that \(ZXY = 60\), which would mean that:

\(180 = (60 - c) + a + 60\)

\(60 + c = a\)

So plugging this result into \(a\) in Quantity A gives us our answer of A.

And you can actually see this is true if you let \(c\) and \(a\) be any of the numbers in the ranges given above.

AS XY=YZ=XZ all the angles are equal
lets consider the point on XZ as w
in the triangle XYW
180=60+a+(60-c)
180-120=a-c
a=60+c
ans: A

Given diagram is an equilateral triangle so all angles are of 60 degree

By the property of exterior angle we can say that
a = c + 60

Therefore QA > QB