Difference between revisions of "1997 AJHSME Problems/Problem 12"
Talkinaway (talk | contribs) (Created page with "==Problem== <math>\angle 1 + \angle 2 = 180^\circ </math> <math>\angle 3 = \angle 4</math> Find <math>\angle 4.</math> <asy> pair H,I,J,K,L; H = (0,0); I = 10*dir(70); J = I ...") |
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label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE); | label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE); | ||
</asy> | </asy> | ||
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<math>\text{(A)}\ 20^\circ \qquad \text{(B)}\ 25^\circ \qquad \text{(C)}\ 30^\circ \qquad \text{(D)}\ 35^\circ \qquad \text{(E)}\ 40^\circ</math> | <math>\text{(A)}\ 20^\circ \qquad \text{(B)}\ 25^\circ \qquad \text{(C)}\ 30^\circ \qquad \text{(D)}\ 35^\circ \qquad \text{(E)}\ 40^\circ</math> | ||
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== See also == | == See also == | ||
− | {{AJHSME box|year=1997|num-b= | + | {{AJHSME box|year=1997|num-b=11|num-a=13}} |
* [[AJHSME]] | * [[AJHSME]] | ||
* [[AJHSME Problems and Solutions]] | * [[AJHSME Problems and Solutions]] | ||
* [[Mathematics competition resources]] | * [[Mathematics competition resources]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:37, 2 May 2021
Problem
Find
Solution
Using the left triangle, we have:
Using the given fact that , we have
.
Finally, using the right triangle, and the fact that , we have:
Thus, the answer is
See also
1997 AJHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.