1985 AHSME Problems/Problem 28
Contents
Problem
In , we have , and . What is ?
Solution 1
Let , so , and thus . Now let be a point on side such that , so , which gives meaning that and are both isosceles, with and . In particular, and . Hence by Stewart's theorem on triangle ,
Solution 2
We apply the law of sines in the form yielding
Now, the angle sum and double angle identities give
Thus our equation becomes Notice, however, that we must have , the latter because otherwise , which would contradict the fact that and are angles in a (non-degenerate) triangle. This means , so the only valid solution is and the fact that is acute also means , so we deduce Accordingly, using the double angle identities again, Finally, the law of sines now gives so, substituting the above results,
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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