In a triangle ABC,right angled at B, AC=b, AB=a. A line is drawn from point C to the opposite side AB intersecting it at D such that AD=DB . Determine sin^2 C+ cos^2C.
 

Asked by Jerlin George | 18th Jun, 2015, 11:02: PM

Expert Answer:

B C equals square root of A C squared minus A B squared end root equals square root of b squared minus a squared end root sin space C equals fraction numerator A B over denominator A C end fraction equals a over b cos space C equals fraction numerator B C over denominator A C end fraction equals fraction numerator square root of b squared minus a squared end root over denominator b end fraction sin squared C plus cos squared C equals open parentheses a over b close parentheses squared plus open parentheses fraction numerator square root of b squared minus a squared end root over denominator b end fraction close parentheses squared equals a squared over b squared plus fraction numerator b squared minus a squared over denominator b squared end fraction equals fraction numerator a squared plus b squared minus a squared over denominator b squared end fraction equals 1 therefore sin squared C plus cos squared C equals 1

Answered by Prasenjit Paul | 19th Jun, 2015, 11:04: AM