Question

In a triangle \[ABC\], if \[\dfrac{{2\cos A}}{a} + \dfrac{{\cos B}}{b} + \dfrac{{2\cos C}}{c} = \dfrac{a}{{bc}} + \dfrac{b}{{ca}}\] . Then what is the value of an angle \[A\]?
A. \[45^ {\circ }\]
B. \[30^ {\circ }\]
C. \[90^ {\circ }\]
D. \[60^ {\circ }\]

Answer 1

Hint: Here, first substitute the values of each angle in the equation by using the laws of cosines. Then, solve the equation in terms of \[a,b\] and \[c\]. Then find the relation between these three sides and use that to find the value of an angle \[A\].

Formula used:
Laws of cosines:
\[\cos A = \dfrac{{{b^2} + c{}^2 - {a^2}}}{{2bc}}\]
\[\cos B = \dfrac{{{c^2} + a{}^2 - {b^2}}}{{2ac}}\]
\[\cos C = \dfrac{{{a^2} + b{}^2 - {c^2}}}{{2ab}}\]

Complete step by step solution:
Given:
In a triangle \[ABC\], \[\dfrac{{2\cos A}}{a} + \dfrac{{\cos B}}{b} + \dfrac{{2\cos C}}{c} = \dfrac{a}{{bc}} + \dfrac{b}{{ca}}\].
Let’s simplify the given equation.
Substitute the value of the angles by using the laws of cosines.
\[\dfrac{{2\left( {\dfrac{{{b^2} + c{}^2 - {a^2}}}{{2bc}}} \right)}}{a} + \dfrac{{\left( {\dfrac{{{c^2} + a{}^2 - {b^2}}}{{2ac}}} \right)}}{b} + \dfrac{{2\left( {\dfrac{{{a^2} + b{}^2 - {c^2}}}{{2ab}}} \right)}}{c} = \dfrac{a}{{bc}} + \dfrac{b}{{ca}}\]
Simplify the left-hand side.
\[\dfrac{{2\left( {{b^2} + c{}^2 - {a^2}} \right)}}{{2abc}} + \dfrac{{{c^2} + a{}^2 - {b^2}}}{{2abc}} + \dfrac{{2\left( {{a^2} + b{}^2 - {c^2}} \right)}}{{2abc}} = \dfrac{a}{{bc}} + \dfrac{b}{{ca}}\]
\[ \Rightarrow \dfrac{{3{b^2} + {c^2} + a{}^2}}{{2abc}} = \dfrac{a}{{bc}} + \dfrac{b}{{ca}}\]
Separate each term on the left-hand side.
\[\dfrac{{3{b^2}}}{{2abc}} + \dfrac{{{c^2}}}{{2abc}} + \dfrac{{a{}^2}}{{2abc}} = \dfrac{a}{{bc}} + \dfrac{b}{{ca}}\]
Multiply the numerator and the denominator of \[\dfrac{a}{{bc}}\] by \[2a\], and \[\dfrac{b}{{ca}}\] by \[2b\].
\[\dfrac{{3{b^2}}}{{2abc}} + \dfrac{{{c^2}}}{{2abc}} + \dfrac{{a{}^2}}{{2abc}} = \dfrac{{2a{}^2}}{{2abc}} + \dfrac{{2{b^2}}}{{2abc}}\]
\[ \Rightarrow \dfrac{{{b^2}}}{{2abc}} + \dfrac{{{c^2}}}{{2abc}} = \dfrac{{a{}^2}}{{2abc}}\]
\[ \Rightarrow {b^2} + {c^2} = a{}^2\]
Since \[a,b\] and \[c\] are the lengths of the sides of the triangle.
Here, the sum of the square of the two sides is equal to the square of the third side.
It follows the converse of the Pythagoras theorem.
Thus, it is a right-angled triangle.
Then, \[a\] will be the hypotenuse of the triangle.
We know that the opposite angle of the hypotenuse is a right angle.
Therefore, the measure of the angle \[A\] is \[90^ {\circ }\].
Hence the correct option is C.

Note: Pythagoras Theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
The converse of the Pythagoras theorem: In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the triangle is a right-angled triangle.