Question

The perimeter of a \[\vartriangle ABC\] is \[6\] times the arithmetic mean of the sines of its angles. If the side a is \[1\] ,then the angle A is
\[\left( A \right){\text{ }}\dfrac{\pi }{6}\]
\[\left( B \right){\text{ }}\dfrac{\pi }{3}\]
\[\left( C \right){\text{ }}\dfrac{\pi }{2}\]
\[\left( D \right){\text{ }}\pi \]

Answer 1

Hint: First we have to use the formula of the perimeter of the triangle. The arithmetic mean of the sine angles is \[\dfrac{{\sin A + \sin B + \sin C}}{3}\] . Then write the equation according to the conditions given in the question. Then apply the sine rule. According to the sine rule, \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R\] . From this you will get the values of a, b and c. On further solving you will get the value of R, on substituting it in the value of a you will get the value of \[\angle A\] .

Complete step by step answer:
Let’s see the diagram first,

It is given to us that the perimeter of a triangle is six times the arithmetic mean of the sine angles. It is also given that the side of the triangle is one.
The perimeter of a triangle is the sum of all three sides of the triangle. Therefore,
Perimeter \[ = \] a \[ + \] b \[ + \] c
We can also write the above equation as
\[2s = a + b + c\] --------- (i)
Because ‘s’ is used in heron’s formula which denotes the semi-perimeter of a triangle, whose area has to be evaluated. Semi-perimeter is equal to the sum of all three sides of the triangle divided by two that is \[s = \dfrac{{a + b + c}}{2}\] where a, b and c are the three sides of the triangle. From which we can say that \[2s = a + b + c\] .
As the value of a is given as one. Therefore, \[2s = 1 + b + c\] --------- (ii)
Also the arithmetic mean of the sine angles will be \[\dfrac{{\sin A + \sin B + \sin C}}{3}\]
According to the question,
\[ \Rightarrow {\text{ }}2s{\text{ }} = {\text{ }}6{\text{ }} \times {\text{ }}\dfrac{{\sin A + \sin B + \sin C}}{3}\]
On dividing six by three we get
\[ \Rightarrow {\text{ }}2s{\text{ }} = {\text{ 2}}\left( {\sin A + \sin B + \sin C} \right)\]
From equation (ii) we have
\[ \Rightarrow {\text{ 1}} + b + c{\text{ }} = {\text{ 2}}\left( {\sin A + \sin B + \sin C} \right)\] ------------- (iii)
Now here we will now apply the sine rule. According to the sine rule \[\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R\] where R is the radius of the circumcircle of ABC. As the value of a is one, therefore
\[ \Rightarrow \dfrac{1}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R\]
Now on further solving we get
\[1 = 2R\sin A\] , \[b = 2R\sin B\] and \[c = 2R\sin C\]
Now put these sine values in equation (iii),
\[ \Rightarrow {\text{ }}2R\sin A + 2R\sin B + 2R\sin C = 2\left( {\sin A + \sin B + \sin C} \right)\]
Take out \[2R\] common in the left hand side
\[ \Rightarrow {\text{ }}2R\left( {\sin A + \sin B + \sin C} \right) = 2\left( {\sin A + \sin B + \sin C} \right)\]
Now the similar terms will cancel out and we are left with
\[ \Rightarrow R = 1\]
By putting this value of R in \[1 = 2R\sin A\] we get
\[ \Rightarrow 1 = 2\left( 1 \right)\sin A\]
On shifting two from the right hand side to the left hand side we get
\[ \Rightarrow \sin A = \dfrac{1}{2}\]
We know that the value of \[\sin 30^\circ \] is \[\dfrac{1}{2}\] .Therefore,
\[ \Rightarrow \sin A = \sin 30^\circ \]
Sine terms on both sides will cancel out and we have
\[ \Rightarrow A = 30^\circ \]
Or \[A = \dfrac{\pi }{6}\]

So, the correct answer is “Option A”.

Note:
Keep in mind the formula of the perimeter of the triangle. Remember that we can denote the perimeter of the triangle by \[2s\] . Also remember the sine rule. Note that sine rule can be used to find an angle from three sides and an angle or a side from three angles and a side. Also remember the trigonometric values of the functions.