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> | <asy> |
Revision as of 11:43, 14 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.
Using the area formula , the area of the triangle is square inches, which is answer choice .
See Also
1960 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
All AHSME Problems and Solutions |