Problem: In this figure β RFS=β FDR,FD=4 inches, DR= 6 inches, FR=5 inches, FS=721β inches. The length of RS, in inches is:
Answer Choices:
A. undetermined
B. 4
C. 521β
D. 6
E. 641β
Solution:
β³RFDβΌβ³RSF (an angle of one triangle equal to an angle of the other triangle and the including sides in proportion).
β΄RFRSβ=RDSFβ,5RSβ=6721ββ,RS=641β
OR
by the law of cosines, 52=42+62β2β
4β
6cosβ D
β΄cosβ D=4827β=cosβ RFS
\therefore RS^{2}=5^{2}+\left(7 \dfrac{1}{2}\right)^{2}-2\left(7 \dfrac{1}{2}\right)(5)\left(\dfrac{27}{48}\right) \quad \therefore \mathrm{RS}=6 \dfrac{1}