# 1999 AMC 8 Problems/Problem 2

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## Problem

What is the degree measure of the smaller angle formed by the hands of a clock at 10 o'clock? $[asy] draw(circle((0,0),2)); dot((0,0)); for(int i = 0; i < 12; ++i) { dot(2*dir(30*i)); } label("3",2*dir(0),W); label("2",2*dir(30),WSW); label("1",2*dir(60),SSW); label("12",2*dir(90),S); label("11",2*dir(120),SSE); label("10",2*dir(150),ESE); label("9",2*dir(180),E); label("8",2*dir(210),ENE); label("7",2*dir(240),NNE); label("6",2*dir(270),N); label("5",2*dir(300),NNW); label("4",2*dir(330),WNW); [/asy]$ $\text{(A)}\ 30 \qquad \text{(B)}\ 45 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 75 \qquad \text{(E)}\ 90$

## Solution

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 $\frac{1}{6}\cdot360^{\circ}=\boxed{60\text{ (C)}}$.

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