Difference between revisions of "1959 AHSME Problems/Problem 36"
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| − | + | Let the triangle be <math>\triangle ABC</math> with <math>\measuredangle BAC=60^{\circ}</math> and <math>AC=80</math>, as in the diagram. Let <math>AB=x</math>. from the problem, we know that <math>BC=90-x</math>. Now, we can apply the [[Law of Cosines]] on <math>\triangle ABC</math> to solve for <math>x</math>: | |
| − | <math>\boxed{\textbf{(D) 17 | + | \begin{align*} |
| + | (90-x)^2 &= 6400+x^2-160\cos(60^{\circ}) \\ | ||
| + | x^2-180x+8100 &= 6400+x^2-80x \\ | ||
| + | -100x &= -1700 \\ | ||
| + | x &= 17 | ||
| + | \end{align*} | ||
| + | Because the problem asks for the shortest side of the triangle and <math>90-x=73>17</math>, our answer is <math>\boxed{\textbf{(D) }17}</math>. | ||
== See also == | == See also == | ||
{{AHSME 50p box|year=1959|num-b=35|num-a=37}} | {{AHSME 50p box|year=1959|num-b=35|num-a=37}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
| + | [[Category:Introductory Geometry Problems]] | ||
Latest revision as of 15:11, 21 July 2024
Problem
The base of a triangle is 
, and one side of the base angle is 
. The sum of the lengths of the other two sides is 
. The shortest side is:
Solution
![[asy] import geometry; size(10cm); point A = (0,0); point B = (17/20,17*sqrt(3)/20); point C = (8,0); triangle ABC = triangle(A,B,C); // Triangle ABC draw(ABC); dot(A); label("A",A,SW); dot(B); label("B",B,NW); dot(C); label("C",C,SE); // Labels markscalefactor = 0.1; draw(anglemark(C,A,B)); label("$60^{\circ}$", A, (1.5,1.5)); label("$80$", midpoint(A--C), S); label("$x$", midpoint(A--B), NW); label("$90-x$", midpoint(B--C), NE); [/asy]](http://latex.artofproblemsolving.com/4/1/5/41565499f5c6d0c9ea191667c4def91b495fadad.png)
Let the triangle be 
with 
and 
, as in the diagram. Let 
. from the problem, we know that 
. Now, we can apply the Law of Cosines on to solve for 
:
\begin{align*}
(90-x)^2 &= 6400+x^2-160\cos(60^{\circ}) \\
x^2-180x+8100 &= 6400+x^2-80x \\
-100x &= -1700 \\
x &= 17
\end{align*}
Because the problem asks for the shortest side of the triangle and 
, our answer is 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 35 |
Followed by Problem 37 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
