Difference between revisions of "1993 AHSME Problems/Problem 23"
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We get: | We get: | ||
− | <math>\frac{AB}{sin(\angle{AXB})} =\frac{AX}{sin(\angle{ABX})}</math> | + | <math>\frac{AB}{\sin(\angle{AXB})} =\frac{AX}{\sin(\angle{ABX})}</math> |
That's equal to | That's equal to | ||
− | <math>\frac{cos(6)}{sin(180-18)} =\frac{AX}{sin(12)}</math> | + | <math>\frac{\cos(6)}{\sin(180-18)} =\frac{AX}{\sin(12)}</math> |
Therefore, our answer is equal to: | Therefore, our answer is equal to: | ||
<math>\fbox{B}</math> | <math>\fbox{B}</math> | ||
− | Note that <math>sin(162) = sin(18)</math>, and don't accidentally put <math>\fbox{C}</math> because you thought 1/ | + | Note that <math>\sin(162) = \sin(18)</math>, and don't accidentally put <math>\fbox{C}</math> because you thought <math>\frac{1}{\sin}</math> was <math>\sec</math>! |
== See also == | == See also == |
Revision as of 00:26, 2 August 2020
Problem
Points and are on a circle of diameter , and is on diameter
If and , then
Solution
We have all the angles we need, but most obviously, we see that right angle in triangle .
Note also that angle is 6 degrees, so length because the diameter, , is 1.
Now, we can concentrate on triangle (after all, now we can decipher all angles easily and use Law of Sines).
We get:
That's equal to
Therefore, our answer is equal to:
Note that , and don't accidentally put because you thought was !
See also
1993 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.