In $$\triangle$$​PQR, 3 $$\angle$$P = 4$$\angle$$​Q = 6$$\angle$$​R, then find the value of ($$\frac{2 \angle P - \angle Q + 3 \angle R}{5}$$).​

Correct option is A

Given:

​In$$\triangle$$​​PQR, 3 $$\angle$$​P = 4$$\angle$$​​Q = 6$$\angle$$​R,  

Concept Used: 

Sum of the all angle of a triangle = 180$$^0$$ 

Solution: 

Let 3 $$\angle$$​P = 4$$\angle$$​​Q = 6$$\angle$$​R,   = 12 unit 

Then ,

$$\angle P = 4$$ unit

$$\angle Q = 3$$ unit

$$\angle R$$ = 2 unit 

Then the value of ($$\frac{2 \angle P - \angle Q + 3 \angle R}{5}$$) 

                          $$= \frac{2 \times 4 - 3 + 3\times 2}{5} \\\ \\= \frac{8 - 3 + 6}{5}\\\ \\ = \frac{11}{5} \ unit$$​

4 unit + 3 unit + 2 unit = $$180^0$$  

9 unit = $$180^0$$ 

1 unit = $$20^0$$

Then the value of $$\frac{11}{5} unit = \frac{11}{5} \times 20 = 44^0$$​