Question

In triangle $ABC$, if $a\cos A + b\cos B + c\cos C = \dfrac{{2\Delta }}{k}$, then $k$ is equal to
(a) $r$
(b) $R$
(c) $s$
(d) ${R^2}$

Answer 1

Here, we need to find the value of $k$. We will rewrite the sine rule with circumradius and use it in the left hand side of the given expression. Then, we will use the formula for sine of a double angle to simplify the expression further. Next, we will use the angle sum property of a triangle and the conditional trigonometric identity to simplify the expression. Finally, we will use the formula for the area of a triangle using two sides and the angle between them to solve the equation and find the value of $k$.

Formula Used:
We will use the following formulas:
The sine rule states that $\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R$, where $a$ is the length of the side opposite to angle $A$, $b$ is the length of the side opposite to angle $B$, $c$ is the length of the side opposite to angle $C$, and $R$ is the circumradius.
The sine of a double angle can be written as $\sin 2x = 2\sin x\cos x$.
If the sum of the angles $A$, $B$, and $C$ is equal to $180^\circ$, then $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$. This is a conditional trigonometric identity.
The area of a triangle can be calculate using two sides and the angle between them, using the formula $\Delta = \dfrac{1}{2}ab\sin C$, where $b$ is the length of the side opposite to angle $B$ and $a$ is the length of the side opposite to angle $A$.

Complete step-by-step answer:
We will use the sine rule with circumradius, and the formula for the area of the triangle involving sine to solve the given problem.

Simplifying the equation $\dfrac{a}{{\sin A}} = \dfrac{b}{{\sin B}} = \dfrac{c}{{\sin C}} = 2R$, we get
$a = 2R\sin A$
$b = 2R\sin B$
$c = 2R\sin C$
Also, we get
$\sin A = \dfrac{a}{{2R}}$
$\sin B = \dfrac{b}{{2R}}$
Now, we will simplify the given equation.
Substituting $a = 2R\sin A$, $b = 2R\sin B$, and $c = 2R\sin C$ in the equation $a\cos A + b\cos B + c\cos C = \dfrac{{2\Delta }}{k}$, we get
$\Rightarrow 2R\sin A\cos A + 2R\sin B\cos B + 2R\sin C\cos C = \dfrac{{2\Delta }}{k}$
Factoring out $R$ from the expressions, we get
$\Rightarrow R\left( {2\sin A\cos A + 2\sin B\cos B + 2\sin C\cos C} \right) = \dfrac{{2\Delta }}{k}$
We know that the sine of a double angle can be written as $\sin 2x = 2\sin x\cos x$.
Therefore, we get
$2\sin A\cos A = \sin 2A$
$2\sin B\cos B = \sin 2B$
$2\sin C\cos C = \sin 2C$
Substituting $2\sin A\cos A = \sin 2A$, $2\sin B\cos B = \sin 2B$, and $2\sin C\cos C = \sin 2C$ in the equation, we get
$\Rightarrow R\left( {\sin 2A + \sin 2B + \sin 2C} \right) = \dfrac{{2\Delta }}{k}$
Now, since $ABC$ is a triangle, therefore, by angle sum property of a triangle, $\angle A + \angle B + \angle C = 180^\circ$.
If the sum of the angles $A$, $B$, and $C$ is equal to $180^\circ$, then $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$. This is a conditional trigonometric identity.
Substituting $\sin 2A + \sin 2B + \sin 2C = 4\sin A\sin B\sin C$ in the equation, we get
$\begin{array}{l} \Rightarrow R\left( {4\sin A\sin B\sin C} \right) = \dfrac{{2\Delta }}{k}\\\ \Rightarrow 4R\sin A\sin B\sin C = \dfrac{{2\Delta }}{k}\end{array}$
Substituting $\sin A = \dfrac{a}{{2R}}$ and $\sin B = \dfrac{b}{{2R}}$ in the equation, we get
$\Rightarrow 4R\left( {\dfrac{a}{{2R}}} \right)\left( {\dfrac{b}{{2R}}} \right)\sin C = \dfrac{{2\Delta }}{k}$
Simplifying the expression, we get
$\Rightarrow \dfrac{{ab}}{R}\sin C = \dfrac{{2\Delta }}{k}$
Multiplying both sides of the equation by $\dfrac{1}{2}$, we get
$\begin{array}{l} \Rightarrow \dfrac{{ab}}{R}\sin C \times \dfrac{1}{2} = \dfrac{{2\Delta }}{k} \times \dfrac{1}{2}\\\ \Rightarrow \dfrac{1}{2}\dfrac{{ab}}{R}\sin C = \dfrac{\Delta }{k}\\\ \Rightarrow \dfrac{1}{R}\left( {\dfrac{1}{2}ab\sin C} \right) = \dfrac{\Delta }{k}\end{array}$
The area of a triangle can be calculate using two sides and the angle between them, using the formula $\Delta = \dfrac{1}{2}ab\sin C$.
Substituting $\dfrac{1}{2}ab\sin C = \Delta$ in the equation, we get
$\begin{array}{l} \Rightarrow \dfrac{1}{R}\left( \Delta \right) = \dfrac{\Delta }{k}\\\ \Rightarrow \dfrac{\Delta }{R} = \dfrac{\Delta }{k}\end{array}$
Simplifying the equation, we get
$\Rightarrow \dfrac{1}{R} = \dfrac{1}{k}$
Therefore, we get the value of $k$ as
$\Rightarrow k = R$
Thus, the value of $k$ is equal to $R$.

$\therefore$ The correct option is option (b).

Note: We used the angle sum property of a triangle to prove that $\angle A + \angle B + \angle C = 180^\circ$. The angle sum property of a triangle states that the sum of the measures of the three interior angles of a triangle is always $180^\circ$. Here, we also need to remember different trigonometric identities so that we can easily solve the question.