Difference between revisions of "1990 AHSME Problems/Problem 14"
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− | An acute isosceles triangle, <math>ABC</math>, is inscribed in a circle. Through <math>B</math> and <math>C</math>, tangents to the circle are drawn, meeting at point <math>D</math>. If <math>\angle{ABC=\angle{ACB}=2\angle{D}</math> and <math>x</math> is the radian measure of <math>\angle{A}</math>, then <math>x=</math> | + | An acute isosceles triangle, <math>ABC</math>, is inscribed in a circle. Through <math>B</math> and <math>C</math>, tangents to the circle are drawn, meeting at point <math>D</math>. If <math>\angle{ABC}=\angle{ACB}=2\angle{D}</math> and <math>x</math> is the radian measure of <math>\angle{A}</math>, then <math>x=</math> |
<math>\text{(A) } \frac{3\pi}{7}\quad | <math>\text{(A) } \frac{3\pi}{7}\quad | ||
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== Solution == | == Solution == | ||
− | + | 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{540}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{540}{7}</math> degrees, the answer is <math>\dfrac{3\pi}{7}</math> which is <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{540}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{540}{7}</math> degrees, the answer is <math>\dfrac{3\pi}{7}</math> | ||
== See also == | == See also == |
Latest revision as of 10:44, 4 February 2016
Problem
An acute isosceles triangle, , is inscribed in a circle. Through and , tangents to the circle are drawn, meeting at point . If 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 which is
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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All AHSME Problems and Solutions |
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