Difference between revisions of "1989 AHSME Problems/Problem 15"
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== Solution 2 (Trig) == | == Solution 2 (Trig) == | ||
− | Using laws of cosines on <math>\bigtriangleup ABC</math> yields <math>49=25+81-2 \cdot 5 \cdot 9 \cdot \cos A</math> | + | Using laws of cosines on <math>\bigtriangleup ABC</math> yields <math>49=25+81-2 \cdot 5 \cdot 9 \cdot \cos A \implies \cos A = \displaystyle\frac{19}{3}.</math> Let <math>AD=x.</math> Using laws of cosines on <math>\bigtriangleup ABD</math> yields <math>25+x^2-2 \cdot 5 \cdot x \cdot \cos A = 25.</math> Fortunately, we know that <math>\cos A = \displaystyle\frac{19}{3}.</math> Plugging this information back into our equation yields <math>x=\displaystyle\frac{19}{3}.</math> Then, we know that <math>DC=9-\displaystyle\frac{19}{3}=\displaystyle\frac{8}{3} \implies \displaystyle\frac{AD}{DC}=\displaystyle\frac{19}{8}.</math> |
== See also == | == See also == |
Revision as of 18:04, 21 June 2020
Contents
Problem
In , , , , and is on with . Find the ratio of .
Solution
Drop the altitude from through , and let be . Then by Pythagoras and after subtracting the first equation from the second, . Therefore the desired ratio is
Solution 2 (Trig)
Using laws of cosines on yields Let Using laws of cosines on yields Fortunately, we know that Plugging this information back into our equation yields Then, we know that
See also
1989 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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All AHSME Problems and Solutions |
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