Difference between revisions of "2014 AMC 10A Problems/Problem 22"
Nachorobot (talk | contribs) (Very simple yet risky method) |
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==Solution 4 (Measuring) == | ==Solution 4 (Measuring) == | ||
− | If we draw rectangle <math>ABCD</math> | + | If we draw rectangle <math>ABCD</math> and whip out a protractor, we can draw a perfect <math>\overline{BE}</math>, almost perfectly <math>15^\circ</math> degrees off of <math>\overline{BC}</math>. Then we can draw <math>\overline{AE}</math>, and use a ruler to measure it. |
We can clearly see that the <math>\overline{AE}</math> is <math>\boxed{\textbf{(E)}~20}</math>. | We can clearly see that the <math>\overline{AE}</math> is <math>\boxed{\textbf{(E)}~20}</math>. | ||
− | NOTE: this method is a last resort, and is pretty risky. Answer choice <math>\textbf{(D)}~11\sqrt{3}</math> is also very close to <math>\textbf{(E)}~20</math>, meaning that we wouldn't be 100% sure of our answer. | + | NOTE: this method is a last resort, and is pretty risky. Answer choice <math>\textbf{(D)}~11\sqrt{3}</math> is also very close to <math>\textbf{(E)}~20</math>, meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of <math>\triangle ADE, we can clearly see that it is a </math>30-60-90<math> triangle, which verifies our answer of </math>20$. |
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2014|ab=A|num-b=21|num-a=23}} | {{AMC10 box|year=2014|ab=A|num-b=21|num-a=23}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:11, 13 January 2018
Contents
Problem
In rectangle , and . Let be a point on such that . What is ?
Solution (Trigonometry)
Note that . (If you do not know the tangent half-angle formula, it is ). Therefore, we have . Since is a triangle,
Solution 2 (Without Trigonometry)
Let be a point on line such that points and are distinct and that . By the angle bisector theorem, . Since is a right triangle, and . Additionally, Now, substituting in the obtained values, we get and . Substituting the first equation into the second yields , so . Because is a triangle, .
Solution 3 (Trigonometry)
By Law of SinesThus,
We see that is a triangle, leaving
Solution 4 (Measuring)
If we draw rectangle and whip out a protractor, we can draw a perfect , almost perfectly degrees off of . Then we can draw , and use a ruler to measure it. We can clearly see that the is .
NOTE: this method is a last resort, and is pretty risky. Answer choice is also very close to , meaning that we wouldn't be 100% sure of our answer. However, If we measure the angles of 30-60-9020$.
See Also
2014 AMC 10A (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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.