1985 AHSME Problems/Problem 28
Contents
Problem
In , we have and . What is ?
Solution 1
From the Law of Sines, we have , or .
We now need to find an identity relating and . We have
.
Thus we have
.
Therefore, or . Notice that we must have because otherwise . We can therefore disregard because then and also we can disregard because then would be in the third or fourth quadrants, much greater than the desired range.
Therefore, , and . Going back to the Law of Sines, we have .
We now need to find .
.
Therefore, .
Solution 2
Let angle be equal to degrees. Then angle is equal to degrees, and angle is equal to degrees. Let be a point on side such that is equal to degrees. Because , angle is equal to degrees. We can now see that triangles and are both isosceles, with and . From isosceles triangle , we now know that , and since , we know that . From isosceles triangle , we now know that . Applying Stewart's Theorem on triangle gives us , which is .
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
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