1956 AHSME Problems/Problem 10: Difference between revisions
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at <math>D</math>. The number of degrees of angle <math>ADB</math> is: | at <math>D</math>. The number of degrees of angle <math>ADB</math> is: | ||
<math>(A) 15 (B) 30 (C) 60 (D) 90 (E) 120</math> | <math>(A) 15 (B) 30 (C) 60 (D) 90 (E) 120</math> | ||
== Solution == | |||
<asy> | <asy> | ||
import olympiad; | import olympiad; | ||
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<math>ABC</math> is an equilateral triangle, so ∠<math>C</math> must be <math>60</math>°. Since <math>D</math> is on the circle and ∠<math>ADB</math> contains arc <math>AB</math>, we know that ∠<math>D</math> is <math>30</math>° <math>\implies \fbox{B}</math>. | <math>ABC</math> is an equilateral triangle, so ∠<math>C</math> must be <math>60</math>°. Since <math>D</math> is on the circle and ∠<math>ADB</math> contains arc <math>AB</math>, we know that ∠<math>D</math> is <math>30</math>° <math>\implies \fbox{B}</math>. | ||
==See Also== | |||
{{AHSME box|year=1956|num-b=9|num-a=11}} | |||
[[Category:Introductory Algebra Problems]] | |||
{{MAA Notice}} | |||
Revision as of 20:30, 12 February 2021
Problem
A circle of radius
inches has its center at the vertex
of an equilateral triangle
and
passes through the other two vertices. The side
extended through
intersects the circle
at
. The number of degrees of angle
is:
Solution
is an equilateral triangle, so ∠
must be
°. Since
is on the circle and ∠
contains arc
, we know that ∠
is
°
.
See Also
| 1956 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 9 |
Followed by Problem 11 | |
| 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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