Given that ABRQ is a rectangle, ABR is a right-angled triangle with the right angle at B.

Welcome back More — been a while! Let's pick up where you left off. a) To find the length of AS, we first need to find the length of AR. Given that ABRQ is a rectangle, ABR is a right-angled triangle with the right angle at B. Step 1: Find AR using the Pythagorean theorem in ABR. AR^2 = AB^2 + BR^2 Given AB = 6 cm and BR = 4 cm. AR^2 = (6 cm)^2 + (4 cm)^2 AR^2 = 36 cm^2 + 16 cm^2 AR^2 = 52 cm^2 AR = sqrt(52) cm = 2sqrt(13) cm Step 2: Find AS using the Pythagorean theorem in ASR. The problem states that plane PQRS is perpendicular to plane ABRQ. Since SR is a side of the rectangle PQRS and SR QR (where QR is the line of intersection of the two planes), SR is perpendicular to the plane ABRQ. Therefore, ASR is a right-angled triangle with the right angle at R. AS^2 = AR^2 + SR^2 We found AR^2 = 52 cm^2 and given RS = 2 cm. AS^2 = 52 cm^2 + (2 cm)^2 AS^2 = 52 cm^2 + 4 cm^2 AS^2 = 56 cm^2 AS = sqrt(56) cm = sqrt(4 × 14) cm AS = 2sqrt(14) cm As a decimal, AS ≈ 7.48 cm (to 2 decimal places). The length of AS is 2sqrt(14) cm b) The angle that AS makes with plane ABRQ is the angle between AS and its projection onto the plane ABRQ. Since SR is perpendicular to plane ABRQ, the projection of S onto the plane ABRQ is R. The projection of A onto the plane ABRQ is A itself. Therefore, the projection of the line AS onto the plane ABRQ is the line AR. The angle AS makes with plane ABRQ is SAR. Step 1: Identify the right-angled triangle. From part (a), we established that ASR is a right-angled triangle at R. Step 2: Use trigonometry to find SAR. In ASR: Opposite side to SAR is SR = 2 cm. Adjacent side to SAR is AR = sqrt(52) cm. ( SAR) = OppositeAdjacent = (SR)/(AR) ( SAR) = (2)/(sqrt(52)) ( SAR) = (2)/(2sqrt(13)) ( SAR) = (1)/(sqrt(13)) SAR = ((1)/(sqrt(13))) SAR ≈ 15.499^ Rounding to one decimal place: SAR ≈ 15.5^ The angle that AS makes with plane ABRQ is 15.5^ Send me the next one 📸