1985 AHSME Problems/Problem 28
Contents
[hide]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 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.