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{ | + | <math>\text{(A) } \frac{2\pi}{7}\quad |
\text{(B) } \frac{4\pi}{9}\quad | \text{(B) } \frac{4\pi}{9}\quad | ||
\text{(C) } \frac{5\pi}{11}\quad | \text{(C) } \frac{5\pi}{11}\quad |
Revision as of 21:20, 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
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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