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}{30}.</math> Now we use law of cosines on <math>\bigtriangleup ABD</math> getting <math>25=25+AD^2-2 \cdot 5 \cdot AD \cos A.</math> Fortunately, we know that <math>\cos A=\displaystyle\frac{19}{3)}</math> so plugging that in our first equation yields <math>AD=\displaystyle\frac{19}{3}.</math> Then, we know <math>DC=\displaystyle\frac{8}{3} \implies \displaystyle\frac{AD}{DC}=\displaystyle\frac{19}{8}.</math> |
== See also == | == See also == |
Revision as of 18:57, 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
Now we use law of cosines on
getting
Fortunately, we know that
so plugging that in our first equation yields
Then, we know
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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