Difference between revisions of "1972 AHSME Problems/Problem 14"
Coolmath34 (talk | contribs) (Created page with "== Problem == A triangle has angles of <math>30^\circ</math> and <math>45^\circ</math>. If the side opposite the <math>45^\circ</math> angle has length <math>8</math>, then...") |
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| + | == Solution 2 == | ||
| + | Use [[Law of Sines]] to get <math>\frac{8}{\frac{\sqrt2}{2}}=\frac{x}{\frac{1}{2}}</math>. The answer is <math>\boxed{\textbf{(B) }4\sqrt{2}}</math>. | ||
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| + | -aopspandy | ||
Latest revision as of 21:28, 22 June 2021
Problem
A triangle has angles of 
and 
. If the side opposite the angle has length 
, then the side opposite the angle has length
Solution
This triangle can be split into smaller 30-60-90 and 45-45-90 triangles. The side opposite the angle has length 
so the 30-60-90 triangle has sides 
and 
One of the legs of the 45-45-90 triangles is 
so the hypotenuse is 
This is also the side opposite the angle, so the answer is 
-edited by coolmath34
Solution 2
Use Law of Sines to get 
. The answer is 
.
-aopspandy
