1969 AHSME Problems/Problem 8
Problem
Triangle is inscribed in a circle. The measure of the non-overlapping minor arcs , and are, respectively, . Then one interior angle of the triangle is:
Solution
Because the triangle is inscribed, the sum of the minor arcs equals . Thus, Solving this yields , so the inscribed angles are , , and . Noting that an angle of is half of its corresponding inscribed angle, so the angles of are , , and .
See also
1969 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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 • 31 • 32 • 33 • 34 • 35 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.