smc

Question

In a narrow alley of width w a ladder of length a is placed with its foot at point P between the walls. Resting against one wall at Q, the distance k above the ground makes a 45^ angle with the ground. Resting against the other wall at R, a distance h above the ground, the ladder makes a 75^ angle with the ground. The width w is equal to

Accepted Answer

FIGURE We know that m QPL=45^ and m RPT=75^. Therefore, m QPR=60^. Because the two ladders are the same length, we know that RP=PQ=a. Since QPR is isosceles with vertex angle 60^, we can conclude that it must be equilateral. Now, since PTR is a right triangle and m TPR=75^, we can conclude that m PRT=15^. Because QPR is equilateral, we know that m QRP = 60^. It then follows that m QRS= m QRP + m PRT = 60^ + 15^ = 75^. Because of ASA, QRS RPT. From there, it follows that QS=TR=h. Since QS is the width of the alley, the answer is h.