Difference between revisions of "1989 AHSME Problems/Problem 15"
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Drop the altitude <math>h</math> from <math>B</math> through <math>AD</math>, and let <math>AD</math> be <math>2x</math>. Then by Pythagoras <cmath>\begin{cases}h^2+x^2=25\\h^2+(9-x)^2=49\end{cases}</cmath> and after subtracting the first equation from the second, <math>x=3\tfrac16</math>. Therefore the desired ratio is <cmath>\frac{6\tfrac13}{2\tfrac23}=\boxed{\frac{19}8}</cmath> | Drop the altitude <math>h</math> from <math>B</math> through <math>AD</math>, and let <math>AD</math> be <math>2x</math>. Then by Pythagoras <cmath>\begin{cases}h^2+x^2=25\\h^2+(9-x)^2=49\end{cases}</cmath> and after subtracting the first equation from the second, <math>x=3\tfrac16</math>. Therefore the desired ratio is <cmath>\frac{6\tfrac13}{2\tfrac23}=\boxed{\frac{19}8}</cmath> | ||
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+ | == Solution 2 (Trig) == | ||
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+ | Using laws of cosines on <math>\bigtriangleup ABC</math> yields <math>49=25+81-2 \cdot 5 \cdot 9 \cdot \cos A</math> | ||
== See also == | == See also == |
Revision as of 18:19, 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
See also
1989 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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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