Difference between revisions of "2022 AMC 12B Problems/Problem 14"
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== Solution == | == Solution == | ||
+ | |||
+ | <math>y=x^2+2x-15</math> intersects the <math>x</math>-axis at points <math>(-5, 0)</math> and <math>(3, 0)</math>. Without loss of generality, let these points be <math>A</math> and <math>C</math> respectively. Also, the graph intersects the y-axis at point <math>B = (0, -15)</math>. | ||
+ | |||
+ | Let the origin be point <math>O = (0, 0)</math>. Note that tiangles <math>AOB</math> and <math>COB</math> are right. | ||
+ | |||
+ | <cmath>\tan(\angle ABC) = \frac{\sin (\angle ABC)}{\cos (\angle ABC)} = \frac{\sin(\angle ABO + \angle OBC)}{\cos(\angle ABO + \angle OBC)}</cmath> | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2022|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2022|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:58, 17 November 2022
Problem
The graph of intersects the
-axis at points
and
and the
-axis at point
. What is
?
Solution
intersects the
-axis at points
and
. Without loss of generality, let these points be
and
respectively. Also, the graph intersects the y-axis at point
.
Let the origin be point . Note that tiangles
and
are right.
See Also
2022 AMC 12B (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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.