1984 AHSME Problems/Problem 27
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Problem
In , is on and is on . Also, , , and . Find .
Solution
Let and . We have by AA, so . Substituting in known values gives , so . Also, , and using the Pythagorean Theorem on , we have , so . Using the Pythagorean Theorem on gives , or . Now, we use the Pythagorean Theorem on to get . Substituting into this gives , or . Simplifying this and moving all of the terms to one side gives , and since , we can divide by to get , from which we find that .
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 26 |
Followed by Problem 28 | |
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