Difference between revisions of "1987 AHSME Problems/Problem 14"
(Created page with "==Problem== <math>ABCD</math> is a square and <math>M</math> and <math>N</math> are the midpoints of <math>BC</math> and <math>CD</math> respectively. Then <math>\sin \theta=</...") |
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\textbf{(C)}\ \frac{\sqrt{10}}{5} \qquad | \textbf{(C)}\ \frac{\sqrt{10}}{5} \qquad | ||
\textbf{(D)}\ \frac{4}{5}\qquad | \textbf{(D)}\ \frac{4}{5}\qquad | ||
− | \textbf{(E)}\ \text{none of these} </math> | + | \textbf{(E)}\ \text{none of these} </math> |
+ | ==Solution== | ||
+ | Use the Sine Area Formula. We can isolate the triangle for which the angle <math>\theta</math> is contained in. WLOG, denote the side length of a triangle as <math>2</math>. Our midpoints are then <math>1</math>. Subtract the areas of the triangles that don't include the area of our desired triangle: <math>4-1-1-\frac{1}{2} = \frac{3}{2}.</math> The Sine Area Formula tells us <math>\frac{1}{2}(\sqrt{5})^2\sin\theta=\frac{3}{2}.</math> Solving this equation, we get <math>\sin\theta=\textbf{(B)}\ \frac{3}{5} \qquad</math> | ||
+ | Solution: Everyoneintexas | ||
== See also == | == See also == |
Latest revision as of 00:32, 15 February 2018
Problem
is a square and
and
are the midpoints of
and
respectively. Then
Solution
Use the Sine Area Formula. We can isolate the triangle for which the angle is contained in. WLOG, denote the side length of a triangle as
. Our midpoints are then
. Subtract the areas of the triangles that don't include the area of our desired triangle:
The Sine Area Formula tells us
Solving this equation, we get
Solution: Everyoneintexas
See also
1987 AHSME (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 • 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.