Difference between revisions of "1959 AHSME Problems/Problem 36"

(created solution page)
 
(diagram)
Line 5: Line 5:
  
 
== Solution ==
 
== 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>
 +
 +
 
<math>\boxed{\textbf{(D) 17}}</math>
 
<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}}

Revision as of 15:04, 21 July 2024

Problem

The base of a triangle is $80$, and one side of the base angle is $60^\circ$. The sum of the lengths of the other two sides is $90$. The shortest side is: $\textbf{(A)}\ 45 \qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 36\qquad\textbf{(D)}\ 17\qquad\textbf{(E)}\ 12$


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]


$\boxed{\textbf{(D) 17}}$

See also

1959 AHSC (ProblemsAnswer KeyResources)
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. AMC logo.png