Question

If in triangle \[ABC\], \[\angle C = {60^o}\] then find the value of \[\dfrac{1}{{a + c}} = \dfrac{1}{{b + c}}\].
A. \[\dfrac{1}{{a + b + c}}\]
B. \[\dfrac{2}{{a + b + c}}\]
C. \[\dfrac{3}{{a + b + c}}\]
D. None of these

Answer 1

Hint:
In the given question, we need to find the value of \[\dfrac{1}{{a + c}} + \dfrac{1}{{b + c}}\], for this we need to use the cosine rule of a triangle. After that we will make an assumption such as \[\dfrac{1}{{a + c}} + \dfrac{1}{{b + c}} = \dfrac{3}{{a + b + c}}\]. After simplification of this assumption, we will compare the cosine rule and the obtained result to find the value of \[\dfrac{1}{{a + c}} + \dfrac{1}{{b + c}}\].


Formula used:
Suppose \[a, b\] and \[c\] are the sides of a triangle \[ABC\] and also \[A,B\] and \[C\] are the angles of a triangle \[ABC\] then by cosine rule,
 \[{c^2} = a + {b^2} - 2ab\left( {\cos C} \right)\].

Complete step by step solution:
We know that \[a, b\] and \[c\] are the sides and \[A, B\]and \[C\] are the angles of a triangle \[ABC\].

According to cosine rule, we can say that
 \[{c^2} = a + {b^2} - 2ab\left( {\cos C} \right)\].
Given that, \[\angle C = {60^o}\]
\[ \Rightarrow {c^2} = a + {b^2} - 2ab\left( {\cos {{60}^o}} \right)\]
\[\therefore {c^2} = a + {b^2} - ab\] …… (1)
Now, consider \[\dfrac{1}{{a + c}} + \dfrac{1}{{b + c}} = \dfrac{3}{{a + b + c}}\]
\[\Rightarrow \dfrac{{b + c + a + c}}{{\left( {a + c} \right)\left( {b + c} \right)}} = \dfrac{3}{{a + b + c}}\]
By simplifying, we get
\[ \Rightarrow \left( {a + b + 2c} \right)\left( {a + b + c} \right) = 3\left( {a + c} \right)\left( {b + c} \right)\]
\[ \Rightarrow {a^2} + ab + ac + ba + {b^2} + bc + 2ac + 2bc + 2{c^2} = 3ab + 3ac + 3bc + 3{c^2}\]
\[ \Rightarrow {a^2} + {b^2} + 2ab - 3ab = 3{c^2} - 2{c^2}\]
By simplifying further, we get
\[ \therefore {a^2} + {b^2} - ab = {c^2}\] ……(2)
By comparing (1) and (2), we get;
\[\dfrac{1}{{a + c}} + \dfrac{1}{{b + c}} = \dfrac{3}{{a + b + c}}\]

Therefore, the correct option is (C).

Additional Information : The Cosine Rule states that “In trigonometry that the square of the length of any side of a given triangle is equal to the sum of the squares of the other sides minus twice the product of the other two sides multiplied by the cosine of angle included between them.”

Note: Many students make mistakes in writing the cosine rule of a triangle. This is the only way through which we can solve the example in the simplest way. Also, it is essential to simplify carefully to get the desired result.