Difference between revisions of "1985 AHSME Problems/Problem 28"
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== Solution 2 == | == Solution 2 == | ||
− | Let angle A be equal to <math>x</math> degrees. Then angle C is equal to <math>3x</math> degrees, and angle B is equal to <math>180-4x</math> degrees. Let D be a point on side AB such that angle ACD is equal to <math>x</math> degrees. Because <math>2x+180-4x+CDB=180</math>, angle CDB is equal to <math>2x</math> degrees. We can now see that triangles CDB and CDA are both isosceles | + | Let angle <math>A</math> be equal to <math>x</math> degrees. Then angle <math>C</math> is equal to <math>3x</math> degrees, and angle <math>B</math> is equal to <math>180-4x</math> degrees. Let <math>D</math> be a point on side <math>AB</math> such that <math>\angle ACD</math> is equal to <math>x</math> degrees. Because <math>2x+180-4x+\angle CDB=180</math>, angle <math>CDB</math> is equal to <math>2x</math> degrees. We can now see that triangles <math>CDB</math> and <math>CDA</math> are both isosceles, with <math>CB=DB</math> and <math>AD=AC</math>. From isosceles triangle <math>CDB</math>, we now know that <math>BD = 27</math>, and since <math>AB = c = 48</math>, we know that <math>AD = 21</math>. From isosceles triangle <math>CDA</math>, we now know that <math>CD = 21</math>. Applying Stewart's Theorem on triangle <math>ABC</math> gives us <math>AC = 35</math>, which is <math>\boxed{\text{B}}</math>. |
==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=27|num-a=29}} | {{AHSME box|year=1985|num-b=27|num-a=29}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:18, 3 April 2018
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 |
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