The diagram shows a play area for babies that is surrounded by partitions at PQ, QR and PS, where PQ = 18 m, PS = 40 m, angle SPQ = angle QRS = 90∘90^\circ and the acute angle PSR = qq can vary.

  1. (i)

    Show that L m, the length of all the partitions can be expressed as

    L=58+40\sinsinθ−18\coscosθ.L = 58 + 40\sinsin \theta - 18\coscos \theta .

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Worked solution

7(i)

1. Set up a geometric construction to find the length of QR.

Draw a perpendicular from PP to SRSR, meeting SRSR at XX. Draw a perpendicular from QQ to PXPX, meeting PXPX at YY.

2. Find the length of PX using triangle PSX.

In right-angled △PSX\triangle PSX, ∠PSX=θ\angle PSX = \theta.
PX=PSsin⁡θ=40sin⁡θPX = PS \sin \theta = 40 \sin \theta

3. Find the length of PY using triangle PQY.

Since PQ⊥PSPQ \perp PS and PX⊥SRPX \perp SR, the angle between PQPQ and PXPX is equal to the angle between PSPS and SRSR, so ∠QPY=θ\angle QPY = \theta.
In right-angled △PQY\triangle PQY,
PY=PQcos⁡θ=18cos⁡θPY = PQ \cos \theta = 18 \cos \theta

4. Express QR in terms of theta and find the total length L.

Since QR⊥SRQR \perp SR and PX⊥SRPX \perp SR, QRQR is parallel to PXPX. With QY⊥PXQY \perp PX, YXRQYXRQ forms a rectangle.
QR=YX=PX−PY=40sin⁡θ−18cos⁡θQR = YX = PX - PY = 40 \sin \theta - 18 \cos \theta
The total length of the partitions is L=PQ+PS+QRL = PQ + PS + QR.
L=18+40+(40sin⁡θ−18cos⁡θ)L = 18 + 40 + (40 \sin \theta - 18 \cos \theta)
L=58+40sin⁡θ−18cos⁡θL = 58 + 40 \sin \theta - 18 \cos \theta (Shown)

Answer

L=58+40sin⁡θ−18cos⁡θL = 58 + 40\sin \theta - 18\cos \theta

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