Difference between revisions of "1963 AHSME Problems/Problem 22"
Rockmanex3 (talk | contribs) (Solution to Problem 22) |
Jasond6789 (talk | contribs) m (Was OBE not BOE.) |
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Acute-angled <math>\triangle ABC</math> is inscribed in a circle with center at <math>O</math>; <math>\stackrel \frown {AB} = 120^\circ</math> and <math>\stackrel \frown {BC} = 72^\circ</math>. | Acute-angled <math>\triangle ABC</math> is inscribed in a circle with center at <math>O</math>; <math>\stackrel \frown {AB} = 120^\circ</math> and <math>\stackrel \frown {BC} = 72^\circ</math>. | ||
− | A point <math>E</math> is taken in minor arc <math>AC</math> such that <math>OE</math> is perpendicular to <math>AC</math>. Then the ratio of the magnitudes of <math>\angle | + | A point <math>E</math> is taken in minor arc <math>AC</math> such that <math>OE</math> is perpendicular to <math>AC</math>. Then the ratio of the magnitudes of <math>\angle BOE</math> and <math>\angle BAC</math> is: |
<math>\textbf{(A)}\ \frac{5}{18}\qquad | <math>\textbf{(A)}\ \frac{5}{18}\qquad |
Revision as of 20:59, 29 April 2023
Problem
Acute-angled is inscribed in a circle with center at
;
and
.
A point is taken in minor arc
such that
is perpendicular to
. Then the ratio of the magnitudes of
and
is:
Solution
Because and
,
. Also,
and
, so
. Since
,
. Finally,
is an isosceles triangle, so
. Because
, the ratio of the magnitudes of
and
is
, which is answer choice
.
See Also
1963 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |
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