In triangle ABC, right-angled at B, if sin A =1213, then find the value of
(sinA cosA + sinC cosC ).
120169
SinA=BCAC =1213
Let BC = 12k and AC = 13k, where k is a positive number.
Using Pythagoras Theorem in Triangle ABC, we get,
AC2=BC2+AB2
⇒(13k)2=(12k)2+(AB)2
⇒AB2=169k2−144k2=25k2
⇒AB = 5k
So, now we can get the required trigonometric ratios:
cos A = ABAC=5k13k=513
sin C = ABAC=5k13k=513
cos C = BCAC=12k13k=1213
⇒ sinA cosA + sin C cos C = 1213×513+513×1213
= 60169+60169=120169