Difference between revisions of "1990 AHSME Problems/Problem 14"
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== Solution == | == Solution == | ||
− | We can make two equations (assume angle D is y): <math>y+2x=180</math> and <math>4y+x=180</math>. We find that <math>x=\dfrac{540}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{540}{7}</math> degrees, the answer is <math>\dfrac{3\pi}{7}</math> which is <math>\fbox{A}</math> | + | We can make two equations (assume angle D is y): <math>y+2x=180</math> and <math>4y+x=180</math>. We find that <math>x=\dfrac{540}{7}</math>. Now we have to convert this to radians. 360 degrees is <math>2\pi</math> radians, so since we have <math>\dfrac{540}{7}</math> degrees, the answer is <math>\dfrac{3\pi}{7}</math> which is <math>\fbox{A}</math> no |
== See also == | == See also == |
Revision as of 13:39, 2 October 2022
Problem
An acute isosceles triangle, , is inscribed in a circle. Through
and
, tangents to the circle are drawn, meeting at point
. If
and
is the radian measure of
, then
Solution
We can make two equations (assume angle D is y): and
. We find that
. Now we have to convert this to radians. 360 degrees is
radians, so since we have
degrees, the answer is
which is
no
See also
1990 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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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