SOME QUESTIONS THAT I'M UNABLE TO SOLVE

Asked by  | 1st May, 2008, 02:49: PM

Expert Answer:

Q 9)     triangle ABC rt angled at B   given tan A = 1 / 3

   tan A =BC / AB = 1 /  3    In triangle ABC  AC = 2 by Pythagores them.

   sin A =1/2 , cos A =  3 / 2 , sin C =  3 / 2 , cos C = 1/2

   sinA cos C + cos A sin C = 1/2 x 1/ 2 +  3 / 2 x  3 / 2

                                                = 1 /4 + 3 / 4 = 1

 cos A cos C - sin A sin C =  3 / 2 x 1/2 - 1 /2 x  3 / 2

                                              =  3 / 4 -  3/ 4 = 0

Q 10 )   I n triangle PQR rt angled at Q PQ = 5 and PR + QR = 25  therefore PR = 25 - QR

  using Pythagores them.     PR 2= QR 2 + PQ2

                                  ( 25 - QR )2 = QR 2 + (5)2

                              625 + QR2 - 50 QR = QR2 + 25

                                  QR = 12 and PR = 13

       sin P =12 / 13 ,cos P = 5 / 13 , tan P =12 / 5

Answered by  | 4th May, 2008, 06:00: PM

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