A room is 12m long, 9m wide and 8m high. Find the cosine of the angle which a diagonal of the room makes with the floor of the room
\(\frac{15}{17}\)
\(\frac{8}{17}\)
\(\frac{8}{15}\)
\(\frac{12}{17}\)
Correct answer is A
ABCD is the floor. By pathagoras
AC\(^2\) = 144 + 81 = \(\sqrt{225}\)
AC = 15cm
Height of room 8m, diagonal of floor is 15m
Therefore, the cosine of the angle which a diagonal of the room makes with the floor is
EC\(^2\) = 15\(^2\) + 8\(^2\) cosine
\(\frac{adj}{Hyp} = \frac{15}{17}\)
EC\(^2\) = \(\sqrt{225 + 64}\)
EC = \(\sqrt{289}\)
= 17