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
An acute isosceles triangle, , is inscribed in a circle. Through and , tangents to the circle are drawn, meeting at point . If $\angle{ABC=\angle{ACB}=2\angle{D}$ (Error compiling LaTeX. ! Missing } inserted.) and is the radian measure of , then
Solution
We can make two equations (assume angle D is y): and . We find that . Now we have to convert this to radians. 360 degrees is radians, so since we have degrees, the answer is .
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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