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

Hint: 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.