H=5; M=51; A=abs((30*H) - (5.5*M));print"A =", A,"Degrees")
A = 130.5 Degrees [This is the smaller angle].
What is the angle between two clock hands at 5:51?
Hello Guest!
\(\omega_{small} =2\pi/12h\\ \alpha_{small}=t\cdot \omega_{small}=(5+\frac{51}{60})h\cdot\frac{\pi}{6h}\)
\(\alpha_{small}=\frac{5+\frac{51}{60}}{6}\pi\)
\(\alpha_{small}=0.975\pi\)
\(\omega_{large} =2\pi/60min\\ \alpha_{large}=51min\cdot \frac{\pi}{30min}\)
\(\alpha_{large}=1.7\pi\)
\(\Delta\alpha=(1.7-0.975)\cdot\pi\\ \Delta\alpha=0.725\cdot\pi\cdot \frac{180\ deg}{\pi}\)
\(\Delta\alpha=130.5\ deg\)
The angle between two clock hands at 5:51 is 130.5 degrees.
!
Hour hand starts at 5 *30 = 150 o moves 1/2 degree per minute 150 + 1/2 x 51 = 175.50
minute hand moves 6o per min 6 x 51 = 306o
306 - 175.5 = 130.5o