Question

In a \[\Delta ABC\], find side b.
A. \[c\cos A + a\cos C\]
B. \[a\cos B + b\cos A\]
c. \[b\cos C + c\cos B\]
D. None of these

Answer 1

Hint: By using the cosine law we will check all options to find which one is the correct formula.

Formula used:
Cosine law:
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]

Complete step by step solution:
First option is \[c\cos A + a\cos C\].
Putting \[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\] and \[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\]in \[c\cos A + a\cos C\]
\[c \cdot \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}} + a \cdot \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\]
Cancel out c and a from first and second term respectively:
\[ = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2b}} + \dfrac{{{a^2} + {b^2} - {c^2}}}{{2b}}\]
\[ = \dfrac{{{b^2} + {c^2} - {a^2} + {a^2} + {b^2} - {c^2}}}{{2b}}\]
\[ = \dfrac{{{b^2} + {b^2}}}{{2b}}\]
\[ = \dfrac{{2{b^2}}}{{2b}}\]
Cancel out 2b from denominator and numerator
\[ = b\]
Second option is \[a\cos B + b\cos A\].
Putting \[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\] and \[\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\]in \[a\cos B + b\cos A\]
\[a \cdot \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}} + b \cdot \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
Cancel out a and b from first and second term respectively:
\[ = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2c}} + \dfrac{{{b^2} + {c^2} - {a^2}}}{{2c}}\]
\[ = \dfrac{{{a^2} + {c^2} - {b^2} + {b^2} + {c^2} - {a^2}}}{{2c}}\]
\[ = \dfrac{{{c^2} + {c^2}}}{{2c}}\]
\[ = \dfrac{{2{c^2}}}{{2c}}\]
Cancel out 2c from denominator and numerator
\[ = c\]
Third option is \[b\cos C + c\cos B\].
Putting \[\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\] and \[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\]in \[b\cos C + c\cos B\]
\[b \cdot \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}} + c \cdot \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\]
Cancel out b and c from first and second term respectively:
\[ = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2a}} + \dfrac{{{a^2} + {c^2} - {b^2}}}{{2a}}\]
\[ = \dfrac{{{a^2} + {b^2} - {c^2} + {a^2} + {c^2} - {b^2}}}{{2a}}\]
\[ = \dfrac{{{a^2} + {a^2}}}{{2c}}\]
\[ = \dfrac{{2{a^2}}}{{2a}}\]
Cancel out 2a from denominator and numerator
\[ = a\]
Hence option A is the correct option.

Note:some students try to solve the given question by solving the cosine formula \[{b^2} = {a^2} + {c^2} - 2ac\cos B\]. But it is not correct way to solve the question. Using the options, we will solve the question.