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

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