1959 AHSME Problems/Problem 36
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
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 | |
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