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