Difference between revisions of "1982 AHSME Problems/Problem 22"
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+ | We know that <math>m\angle QPL=45^{\circ}</math> and <math>m\angle RPT=75^{\circ}.</math> Therefore, <math>m\angle QPR=60^{\circ}.</math> <cmath>\qquad</cmath> | ||
+ | Because the two ladders are the same length, we know that | ||
+ | <cmath>RP=PQ=a.</cmath> | ||
+ | Since <math>\triangle QPR</math> is isosceles with vertex angle <math>60^{\circ},</math> we can conclude that it must be equilateral.<cmath>\qquad</cmath> | ||
+ | Now, since <math>\triangle PTR</math> is a right triangle and <math>m\angle TPR=75^{\circ},</math> we can conclude that <math>m\angle PRT=15^{\circ}.</math> It then follows that <math>m\angle QRS=75^{\circ}.</math> <cmath>\qquad</cmath> | ||
+ | Because of ASA, <math>\triangle QRS\cong\triangle RPT.</math> From there, it follows that <math>QS=TR=h.</math> Since <math>QS</math> is the width of the alley, the answer is <math>\boxed{\text{E) }h}.</math> | ||
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+ | ~ Saumya Singhal |
Revision as of 22:45, 24 December 2020
Problem 22
In a narrow alley of width a ladder of length is placed with its foot at point P between the walls. Resting against one wall at , the distance k above the ground makes a angle with the ground. Resting against the other wall at , a distance h above the ground, the ladder makes a angle with the ground. The width is equal to
Solution
We know that and Therefore, Because the two ladders are the same length, we know that Since is isosceles with vertex angle we can conclude that it must be equilateral. Now, since is a right triangle and we can conclude that It then follows that Because of ASA, From there, it follows that Since is the width of the alley, the answer is
~ Saumya Singhal