Difference between revisions of "1966 AHSME Problems/Problem 6"

Latest revision as of 11:42, 28 January 2020

Problem

$AB$ is the diameter of a circle centered at $O$ . $C$ is a point on the circle such that angle $BOC$ is $60^\circ$ . If the diameter of the circle is $5$ inches, the length of chord $AC$ , expressed in inches, is:

$\text{(A)} \ 3 \qquad \text{(B)} \ \frac {5\sqrt {2}}{2} \qquad \text{(C)} \frac {5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$

Solution

$[asy] draw(unitcircle); draw((-1,0)--(1,0)--(1/2, sqrt(3)/2)--cycle); label( "A", (-1,0), W); label( "B", (1,0), E); label( "C", (1/2, sqrt(3)/2), N); [/asy]$ We see that $\angle A$ is half the measure of $\angle BOC$ , so $\angle A = 30^{\circ}$ . That makes $ABC$ a $30-60-90$ triangle and sidelength $\overline{AC}$ equal to $\frac{5\sqrt{3}}{2}$ . $\fbox{C}$

1966 AHSME (Problems • Answer Key • Resources)
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Difference between revisions of "1966 AHSME Problems/Problem 6"

Latest revision as of 11:42, 28 January 2020

Problem

Solution

See also

@@ Line 5: / Line 5: @@
 == Solution ==
+<asy>
+draw(unitcircle);
+draw((-1,0)--(1,0)--(1/2, sqrt(3)/2)--cycle);
+label( "A", (-1,0), W);
+label( "B", (1,0), E);
+label( "C", (1/2, sqrt(3)/2), N);
+</asy>
+We see that <math>\angle A</math> is half the measure of <math>\angle BOC</math>, so <math>\angle A = 30^{\circ}</math>. That makes <math>ABC</math> a <math>30-60-90</math> triangle and sidelength <math>\overline{AC}</math> equal to <math>\frac{5\sqrt{3}}{2}</math>.
 <math>\fbox{C}</math>