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Difference between revisions of "1964 AHSME Problems/Problem 35"

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<math>\textbf{(A) }3:11\qquad\textbf{(B) }5:11\qquad\textbf{(C) }1:2\qquad\textbf{(D) }2:3\qquad \textbf{(E) }25:33</math>
 
<math>\textbf{(A) }3:11\qquad\textbf{(B) }5:11\qquad\textbf{(C) }1:2\qquad\textbf{(D) }2:3\qquad \textbf{(E) }25:33</math>
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==Solution==
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Using Law of Cosines
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and the fact that the ratio equals cos(a)/[cos(b)cos(c)]
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B 5:11
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==See Also==
 
==See Also==
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{{MAA Notice}}
 
{{MAA Notice}}
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Using Law of Cosines
 
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and the fact that the ratio equals cos(a)/[cos(b)cos(c)]
 
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B 5:11
 

Revision as of 19:21, 18 April 2020

Problem

The sides of a triangle are of lengths $13$, $14$, and $15$. The altitudes of the triangle meet at point $H$. if $AD$ is the altitude to the side of length $14$, the ratio $HD:HA$ is:

$\textbf{(A) }3:11\qquad\textbf{(B) }5:11\qquad\textbf{(C) }1:2\qquad\textbf{(D) }2:3\qquad \textbf{(E) }25:33$

Solution

Using Law of Cosines and the fact that the ratio equals cos(a)/[cos(b)cos(c)] B 5:11


See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 34 		Followed by
Problem 36
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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