In a \[\vartriangle ABC\] , if \[2\angle A = 3\angle B = 6\angle C\] , calculate \[\angle A\] , \[\angle B\] and \[\angle C\] .

Hint: In the above given question, we are given a triangle \[\vartriangle ABC\] in which the three angles \[\angle A\] , \[\angle B\] and \[\angle C\] are given as such that two times of \[\angle A\] is equal to three times of \[\angle B\] which is then equal to six times of \[\angle C\] . We have to calculate the values of the measure of all the three angles \[\angle A\] , \[\angle B\] and \[\angle C\] . In order to approach the required solution, we have to consider the property of a triangle known as the angle sum property of a triangle.

Complete answer:
Given that, a triangle \[\vartriangle ABC\] such that,
\[ \Rightarrow 2\angle A = 3\angle B = 6\angle C\]
We have to find the value of the three angles \[\angle A\] , \[\angle B\] and \[\angle C\] .
Now since it is given that \[2\angle A = 3\angle B = 6\angle C\]
Hence, we can also write is as,
\[ \Rightarrow 2\angle A = 6\angle C\]
That is,
\[ \Rightarrow \angle A = 3\angle C\] ...(1)
Similarly, we can write \[2\angle A = 3\angle B = 6\angle C\] as,
\[ \Rightarrow 3\angle B = 6\angle C\]
That is,
\[ \Rightarrow \angle B = 2\angle C\] ...(2)
Now, since we know that from the angle sum property of the triangle, the sum of all three angles in a triangle is equal to \[180^\circ \] .
Therefore, applying the angle sum property of a triangle in the given triangle \[\vartriangle ABC\] , we can write
\[ \Rightarrow \angle A + \angle B + \angle C = 180^\circ \]
Substituting the values from equations 1 and 2 in this equation, we get
\[ \Rightarrow 3\angle C + 2\angle C + \angle C = 180^\circ \]
That gives us,
\[ \Rightarrow 6\angle C = 180^\circ \]
Hence,
\[ \Rightarrow \angle C = 30^\circ \]
Therefore,
\[ \Rightarrow \angle A = 3\angle C = 3 \times 30^\circ \]
That is,
\[ \Rightarrow \angle A = 90^\circ \]
And,
\[ \Rightarrow \angle B = 2\angle C = 2 \times 30^\circ \]
That is,
\[ \Rightarrow \angle B = 60^\circ \]
Therefore, the three angles of \[\vartriangle ABC\] are \[\angle A = 90^\circ \] , \[\angle B = 60^\circ \] and \[\angle C = 30^\circ \] .

Note:
We can note that in the above given triangle \[\vartriangle ABC\] , we have \[\angle A = 90^\circ \] . Now, an angle of \[\vartriangle ABC\] is a right angle, therefore the above given triangle \[\vartriangle ABC\] is a right angled triangle. Hence, the side opposite to the right angle \[\angle A = 90^\circ \] , that is the side BC, is the hypotenuse for the triangle \[\vartriangle ABC\] while the other two sides AB and AC are the perpendicular sides.