
For \[\Delta ABC\], find \[\dfrac{{\cos A}}{a} + \dfrac{{\cos B}}{b} + \dfrac{{\cos C}}{c}\].
A. \[\dfrac{{{a^2} + {b^2} + {c^2}}}{{abc}}\]
B. \[\dfrac{{{a^2} + {b^2} + {c^2}}}{{2abc}}\]
C. \[\dfrac{{2\left( {{a^2} + {b^2} + {c^2}} \right)}}{{abc}}\]
D. \[{a^2} + {b^2} + {c^2}\]
Answer
160.8k+ views
Hint: First we will find \[\cos A\], \[\cos B\], \[\cos C\] from the cosine law. Then we will substitute the value of \[\cos A\], \[\cos B\], \[\cos C\] in the given expression.
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:
We know the cosine law
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]
Rewrite the cosine law
\[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
\[\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\]
\[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\]
Now substitute \[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\], \[\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\] and \[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\] in the given expression:
\[\dfrac{{\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{a} + \dfrac{{\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}}}{b} + \dfrac{{\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}}}{c}\]
\[ = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2abc}} + \dfrac{{{a^2} + {c^2} - {b^2}}}{{2abc}} + \dfrac{{{a^2} + {b^2} - {c^2}}}{{2abc}}\]
\[ = \dfrac{{{b^2} + {c^2} - {a^2} + {a^2} + {c^2} - {b^2} + {a^2} + {b^2} - {c^2}}}{{2abc}}\]
\[ = \dfrac{{{a^2} + {b^2} + {c^2}}}{{2abc}}\]
Hence option B is the correct option.
Additional information:
The sine laws and cosine laws are based on oblique triangles. Sine is applicable on the problem in which we know at least 2 sides and 1 opposite angle any one of the sides or 2 angles and 1 opposite side one of the angles.
Note: Students often make mistakes with sine formula and cosine formula. They take \[\dfrac{{\cos A}}{a} = \dfrac{{\cos B}}{b} = \dfrac{{\cos C}}{c}\] as sine which is an incorrect formula. Using the incorrect formula, they try to solve the given question. The correct formula for sine formula is \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]. This question is based on the cosine 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:
We know the cosine law
\[{a^2} = {b^2} + {c^2} - 2bc\cos A\]
\[{b^2} = {a^2} + {c^2} - 2ac\cos B\]
\[{c^2} = {a^2} + {b^2} - 2ab\cos C\]
Rewrite the cosine law
\[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\]
\[\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\]
\[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\]
Now substitute \[\cos A = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\], \[\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}\] and \[\cos C = \dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\] in the given expression:
\[\dfrac{{\dfrac{{{b^2} + {c^2} - {a^2}}}{{2bc}}}}{a} + \dfrac{{\dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}}}{b} + \dfrac{{\dfrac{{{a^2} + {b^2} - {c^2}}}{{2ab}}}}{c}\]
\[ = \dfrac{{{b^2} + {c^2} - {a^2}}}{{2abc}} + \dfrac{{{a^2} + {c^2} - {b^2}}}{{2abc}} + \dfrac{{{a^2} + {b^2} - {c^2}}}{{2abc}}\]
\[ = \dfrac{{{b^2} + {c^2} - {a^2} + {a^2} + {c^2} - {b^2} + {a^2} + {b^2} - {c^2}}}{{2abc}}\]
\[ = \dfrac{{{a^2} + {b^2} + {c^2}}}{{2abc}}\]
Hence option B is the correct option.
Additional information:
The sine laws and cosine laws are based on oblique triangles. Sine is applicable on the problem in which we know at least 2 sides and 1 opposite angle any one of the sides or 2 angles and 1 opposite side one of the angles.
Note: Students often make mistakes with sine formula and cosine formula. They take \[\dfrac{{\cos A}}{a} = \dfrac{{\cos B}}{b} = \dfrac{{\cos C}}{c}\] as sine which is an incorrect formula. Using the incorrect formula, they try to solve the given question. The correct formula for sine formula is \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]. This question is based on the cosine formula.
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