A lvl H2 Maths: Trigonometry

The diagram shows the frame of a writing desk. AB, DC, PQ, SR are perpendicular to the horizontal base BQRC. The rectangle ADSP represents the sloping plane of the desk. ABQP and DCRS are trapezia with SR = PQ = 30 cm, DC = AB = 6 cm, CR = BQ = 42 cm, and BC, AD, QR, PS are each 56 cm.

Giving each answer correct to the nearest 0.1°, find

(i) the angle between PC and plane PQRS

(ii) the angle between the lines RB and SA

(iii) the angle between the planes ADSP and PSXY where X, Y are points on CR and BQ respectively with CX = BY = 12 cm

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Answer:

(i) Angle between PC and plane PQRS is angle CPR

Using pythagoras theorem,
PR² = PR² + QR² = 30² + 56²
PR = 63.5 cm

So, angle CPR = tan^-1 42/63.5 = 33.5°


(ii) Angle between the lines RB and SA = angle SAR' as shown in the diagram below

SR' = SR - DC = 30 cm - 6 cm = 24 cm
AR' = BR = √(BC² + CR²) = √(56² + 42²) = 70 cm

Thus, angle SAR' = tan^-1 24/70 = 18.9°


(iii) Angle between the planes ADSP and PSXY is angle DSX

XR = CR - CX = 42 cm - 12 cm = 30 cm
Therefore, ΔXSR is a isosceles triangle. Since angle SRX = 90°, angle XSR = 45°.

DR' = CR = 42 cm
SR' = 24 cm
Angle DSR' = tan^-1 DR' / SR' = tan^-1 42/24 = 60.255°

angle DSX = angle DSR' - angle XSR' = 69.266° - 45° = 15.3°

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