Question

In a triangle ABC if 2∠A=3∠B=6∠C, then find the value of \[\dfrac{{\angle {\text{A + }}\angle {\text{B}}}}{{\angle {\text{C}}}} \times \angle {\text{B}}\].

Answer 1

- HINT- Proceed the solution of this question by assuming the given relation among angles of triangle equal to a certain variable and apply theorem i.e. sum of all the angles of triangle is 180°.

Complete step-by-step solution -

In ΔABC,
Let all the given relations among angles of ΔABC equal to x.
2∠A=3∠B=6∠C= x
So,
$ \Rightarrow \angle {\text{A = }}\dfrac{{\text{x}}}{2}$
$ \Rightarrow \angle {\text{B = }}\dfrac{{\text{x}}}{3}$
$ \Rightarrow \angle {\text{C = }}\dfrac{{\text{x}}}{6}$​

We know that
Sum of all the angles of triangle is 180°
So on equalising sum of all angles to 180°

Therefore, $\angle {\text{A + }}\angle {\text{B + }}\angle {\text{C = 18}}{{\text{0}}^0}$
On putting the values of all angles
\[ \Rightarrow {\text{ }}\dfrac{{\text{x}}}{2} + \dfrac{{\text{x}}}{3} + \dfrac{{\text{x}}}{6} = {180^0}\]
On further solving
\[ \Rightarrow {\text{ }}\dfrac{{{\text{3x + 2x + x}}}}{6} = {\text{x = }}{180^0}\]
\[ \Rightarrow {\text{x = }}{180^0}\]
On putting the values of x in different angles
$ \Rightarrow \angle {\text{A = }}\dfrac{{{{180}^0}}}{2} = {90^0}$
$ \Rightarrow \angle {\text{B = }}\dfrac{{{{180}^0}}}{3} = {60^0}$
$ \Rightarrow \angle {\text{C = }}\dfrac{{{{180}^0}}}{6} = {30^0}$
In the given question, is asked the value of expression

\[ \Rightarrow \dfrac{{\angle {\text{A + }}\angle {\text{B}}}}{{\angle {\text{C}}}} \times \angle {\text{B = }}\dfrac{{{\text{9}}{{\text{0}}^0} + 6{{\text{0}}^0}}}{{{{30}^0}}} \times {60^0}\]
\[ \Rightarrow \dfrac{{\angle {\text{A + }}\angle {\text{B}}}}{{\angle {\text{C}}}} \times \angle {\text{B }} = {150^0} \times {20^0} = {3000^0}\]

Note- In this particular question, we should know that in a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn). A triangle has three angles, one at each vertex, bounded by a pair of adjacent sides.
We can also see this thing as we know that the sum of all internal angles of a polygon is $\left( {{\text{n - 2}}} \right) \times {180^0}$ where n is the number of sides. As for triangle n = 3
Therefore, $\left( {{\text{3 - 2}}} \right) \times {180^0} = {180^0}$ i.e. sum of all the angles of the triangle is 180°.