Difference between revisions of "1990 AHSME Problems/Problem 14"

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== Solution ==
 
== Solution ==
 
<math>\fbox{A}</math>
 
<math>\fbox{A}</math>
 +
We can make two equations (assume angle D is y): <math>y+2x=180</math> and <math>4y+x=180</math>. We find that <math>x=\dfrac{360}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{360}{7}</math> degrees, the answer is <math>\dfrac{2\pi}{7}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 21:24, 20 November 2014

Problem

[asy] draw(circle((0,0),1),black); draw((0,1)--(cos(pi/14),-sin(pi/14))--(-cos(pi/14),-sin(pi/14))--cycle,dot); draw((-cos(pi/14),-sin(pi/14))--(0,-1/cos(3pi/7))--(cos(pi/14),-sin(pi/14)),dot); draw(arc((0,1),.25,230,310)); MP("A",(0,1),N);MP("B",(cos(pi/14),-sin(pi/14)),E);MP("C",(-cos(pi/14),-sin(pi/14)),W);MP("D",(0,-1/cos(3pi/7)),S); MP("x",(0,.8),S); [/asy]

An acute isosceles triangle, $ABC$, is inscribed in a circle. Through $B$ and $C$, tangents to the circle are drawn, meeting at point $D$. If $\angle{ABC=\angle{ACB}=2\angle{D}$ (Error compiling LaTeX. ! Missing } inserted.) and $x$ is the radian measure of $\angle{A}$, then $x=$

$\text{(A) } \frac{2\pi}{7}\quad \text{(B) } \frac{4\pi}{9}\quad \text{(C) } \frac{5\pi}{11}\quad \text{(D) } \frac{6\pi}{13}\quad \text{(E) } \frac{7\pi}{15}$

Solution

$\fbox{A}$ We can make two equations (assume angle D is y): $y+2x=180$ and $4y+x=180$. We find that $x=\dfrac{360}{7}$. Now we have to convert this to radians. 360 degrees is $2\pi$ radians, so since we have $\dfrac{360}{7}$ degrees, the answer is $\dfrac{2\pi}{7}$.

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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