Difference between revisions of "1960 AHSME Problems/Problem 4"
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If two of the angles are <math>60^{\circ}</math>, then the other angle is <math>60^{\circ}</math> because angles in triangle add up to <math>180^{\circ}</math>. That makes the triangle an equilateral triangle, so all sides are <math>4</math> inches long. | If two of the angles are <math>60^{\circ}</math>, then the other angle is <math>60^{\circ}</math> because angles in triangle add up to <math>180^{\circ}</math>. That makes the triangle an equilateral triangle, so all sides are <math>4</math> inches long. | ||
| − | + | <asy> | |
draw((0,0)--(50,0)--(25,43.301)--cycle); | draw((0,0)--(50,0)--(25,43.301)--cycle); | ||
label("$4$",(10,25)); | label("$4$",(10,25)); | ||
Revision as of 10:55, 8 May 2018
Problem
Each of two angles of a triangle is 
and the included side is 
inches. The area of the triangle, in square inches, is:
Solution
If two of the angles are , then the other angle is because angles in triangle add up to 
. That makes the triangle an equilateral triangle, so all sides are inches long.
![[asy] draw((0,0)--(50,0)--(25,43.301)--cycle); label("$4$",(10,25)); label("$2$",(12.5,-5)); label("$2$",(37.5,-5)); label("$4$",(40,25)); draw((25,43.301)--(25,0)); label("$2\sqrt{3}$",(20,15)); draw((25,3)--(28,3)--(28,0)); [/asy]](http://latex.artofproblemsolving.com/2/6/1/261d9e8db43b79e4fd7a011bc2750389a8bcec98.png)
Using the area formula 
, the area of the triangle is 
square inches, which is answer choice 
.
See Also
| 1960 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 3 |
Followed by Problem 5 | |
| 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 | ||
