Principle Of Mathematical Induction

Question
CBSEENMA11014342

In any triangle ABC, prove that

straight a over sinA space equals space straight b over sinB space equals space fraction numerator straight c over denominator sin space straight C end fraction comma

i.e. sides of the triangle are proportional to the sines of the opposite angles.

Solution

Proof:
                 
Let angle C be acute in figure (i), obtuse in figure (ii) and the right angle in figure (iii)
In each triangle, draw AD perpendicular on BC i.e. AD perpendicular BC (on produced if necessary)
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or                            AD = c sinB                                                              ...(i)
                                                                                                  [∵ AB = C]
In figure (i),           AD over AC space equals space sinC space or space AD space equals space AC space sinC
or                           AD = b sinC                                                   [∵ AC = b]
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or                             AD = b sinC
In figure (iii),             AD over AC space equals space 1 space equals space sin space 90 degree space equals space sin space straight C
or                               AD = b sinC
In all figures,                AD = b sin C                                                         ...(ii)
By (i) and (ii), we get
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Similarly,              straight a over sinA space equals space straight b over sinB space or space fraction numerator straight a over denominator sin space straight A end fraction space equals space fraction numerator straight c over denominator sin space straight C end fraction
Hence,                   space straight a over sinA space equals space straight b over sinB space equals space straight c over sinC
 

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