Question

Let U be the set of all triangles in a plane. If A is the set of all triangles with at least one angle different from $60^\circ $,what is $A'$?

Answer 1

Hint: We have the universal set as the set of all triangles in a plane. A is the set of all triangles with at least one angle different from $60^\circ $. We can compare the condition of angles with the equilateral triangle.

Complete step-by-step answer:
We know that triangles are polygons with 3 sides and 3 angles. Equilateral triangles are triangles that have all the sides and angles are equal. The measures of all the angles will be equal to $60^\circ $.
It is given that A is the set of all triangles in a plane with at least one angle different from $60^\circ $. We know that if one of the angles is not $60^\circ $ it will not be an equilateral triangle. Therefore, A is the set of triangles that are not equilateral.
Complement of a set is defined as the set of elements in the universal set that are not included in the original set.
$A'$ is the complement of A. Here the universal set is the set of all triangles in a plane. So, the complement of A is the set of all the triangles that are not non-equilateral. That means the complement of A is the set of equilateral triangles.
Therefore, $A'$ is the set of all triangles in a plane which are equilateral.

Note: Alternate approach to the problem is,
A is the set of all the triangles with at least one angle is different from $60^\circ $
Then,$A'$ is the set of all the triangles with no angle that is different from $60^\circ $.
Therefore, $A'$ is the set of triangles with all angles are $60^\circ $.
We know that the triangles with all angles $60^\circ $ is an equilateral triangle. So, $A'$ is the set of all equilateral triangles in a plane.
The concept of set theory is used to solve this problem. Complement of a set A is defined as the set of elements in the universal set that are not in the set A.