Difference between revisions of "2014 AMC 10A Problems/Problem 22"
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==Solution 4 (Measuring) == | ==Solution 4 (Measuring) == | ||
− | 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> | + | 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> 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>. | ||
Revision as of 02:53, 15 February 2018
Contents
[hide]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
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
, we can clearly see that it is a
triangle, which verifies our answer of
.
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.