1999 AMC 8 Problems/Problem 2

Problem

What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock?

$[asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); } label("3",2*dir(0),W); label("2",2*dir(30),WSW); label("1",2*dir(60),SSW); label("12",2*dir(90),S); label("11",2*dir(120),SSE); label("10",2*dir(150),ESE); label("9",2*dir(180),E); label("8",2*dir(210),ENE); label("7",2*dir(240),NNE); label("6",2*dir(270),N); label("5",2*dir(300),NNW); label("4",2*dir(330),WNW); [/asy]$

$\textbf{(A)}\ 30 \qquad \textbf{(B)}\ 45 \qquad \textbf{(C)}\ 60 \qquad \textbf{(D)}\ 75 \qquad \textbf{(E)}\ 90$

Solution 1

At $10:00$, the hour hand will be on the $10$ while the minute hand on the $12$.

This makes them $\frac{1}{6}$th of a circle apart, and $\frac{1}{6}\cdot360^{\circ}=\boxed{\textbf{(C) } 60}$.

Solution 2

We know that the full clock is a circle, and therefore has 360 degrees. Considering that there are $12$ numbers, the distance between one number will be $360\div 12=30$. If the time is $10:00$, then the hour hand will be on $10$, and the minute hand will be on, $12$, making them $2$ numbers apart, so they will be $60$ degrees apart, or $\boxed{\textbf{(C) } 60}$