In \(\bigtriangleup\)PQR, PQ = 10cm, QR = 8cm and RP = 6cm, the perpendicular RS is drawn from R to PQ. Find the length of RS

A.

4cm

B.

32cm

C.

\(\frac{30}{7}\)

D.

\(\frac{40}{7}\)

E.

4.8cm

Correct answer is E

Cos Q = \(\frac{r^2 + p^2 - q^2}{2rp}\)

= \(\frac{10^2 + 8^2 - 6^2}{2(10)(8)}\)

= \(\frac{164 - 36}{160}\)

= \(\frac{128}{160}\)

= 0.8

Q = Cos-1 o.8

= 37o x rep. from rt< RSQ, Let RS = x

\(\frac{x}{sin 37^o}\) = \(\frac{8}{sin 90^o}\)

but sin 90o = 1

x = 8 sin 37o

x = 4.8cm