Question

In $\Delta ABC,\;{\text{if}}\;\dfrac{b}{{{c^2} - {a^2}}} + \dfrac{a}{{{c^2} - {b^2}}} = 0,$ then
A) $A = {60^0}$
B) $B = {60^0}$
C) $C = {60^0}$
D) $C = {90^0}$

Answer 1

Hint: To find which of the given options is correct, simplify the given condition by taking LCM (lowest common factor) and further doing algebraic operations. Then use ${a^3} + {b^3} = (a + b)\left( {{a^2} - ab + {b^2}} \right)$ to simplify the equation further and at last use the cosine formula for a given triangle to find the correct option. Cosine formula is given as
In a triangle $ABC$,
$\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$
Use this formula to find the correct option.

Formula used:
Sum of cube of two numbers: ${a^3} + {b^3} = (a + b)\left( {{a^2} - ab + {b^2}} \right)$
Formula for cosine of an angle in a triangle: $\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}$

Complete step-by-step solution:
In order to find the correct option, we will go with the given condition and simplify it as follows
$ \Rightarrow \dfrac{b}{{{c^2} - {a^2}}} + \dfrac{a}{{{c^2} - {b^2}}} = 0$
Taking LCM, we will get
\[
   \Rightarrow \dfrac{{b\left( {{c^2} - {b^2}} \right) + a\left( {{c^2} - {a^2}} \right)}}{{\left( {{c^2} - {a^2}} \right)\left( {{c^2} - {b^2}} \right)}} = 0 \\
   \Rightarrow b\left( {{c^2} - {b^2}} \right) + a\left( {{c^2} - {a^2}} \right) = 0 \\
   \Rightarrow b{c^2} - {b^3} + a{c^2} - {a^3} = 0 \\
   \Rightarrow b{c^2} + a{c^2} = {a^3} + {b^3} \\
   \Rightarrow {c^2}(b + a) = {a^3} + {b^3} \\
 \]
Now, using the formula of addition of cube of two numbers that is given as ${a^3} + {b^3} = (a + b)\left( {{a^2} - ab + {b^2}} \right)$ using this formula, we will get
\[
   \Rightarrow {c^2}(b + a) = (a + b)\left( {{a^2} - ab + {b^2}} \right) \\
   \Rightarrow {c^2} = \left( {{a^2} - ab + {b^2}} \right) \\
 \]
Now we can further write it as
\[ \Rightarrow {a^2} + {b^2} - {c^2} = ab\]
From the formula for cosine of an angle in a triangle, we know that in a triangle $ABC$
$\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}},\;{\text{where}}\;a,\;b,\;c\;{\text{and}}\;C$ are sides of the triangle and angle of the triangle respectively.
Using this, we can write the equation further as
\[
   \Rightarrow 2ab\cos C = ab \\
   \Rightarrow 2\cos C = 1 \\
   \Rightarrow \cos C = \dfrac{1}{2} \\
 \]
And from the trigonometric values table, we know that value of cosine function equals half at an angle of ${60^0}$
That means, $C = {60^0}$

Therefore the correct answer is option ‘C’.

Note: Take in note that we have considered that the side and angle pairs of $(a,\;A),\;(b,\;B)\;{\text{and}}\;(c,\;C)$ are opposite to each other that is in a pair the side is present opposite to the angle it has in its pair. Also we have cancelled $(b + a)\;{\text{and}}\;(a + b)$ from each other because addition holds true for commutative property and thus both are the same.