A landscape artist plans to plant flowers within two concentric circles around a vertical light pole PQ. R is a point on the inner circle and S is a point on the outer circle. R, Q and S lie in the same horizontal plane. RS is a pipe used for the irrigation system in the garden.
The radius of the inner circle is r units and the radius of the outer circle is QS.
The angle of elevation from S to P is 30∘.
RQ^S=2x and PQ=3r
7.1
Show that QS=3r
(3)
7.2
Determine, in terms of r, the area of the flower garden.
(2)
7.3
Show that RS=r10−6cos2x
(3)
7.4
If r=10 metres and x=56∘, calculate RS.
(2)
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- Paper 2
Question 7
The diagram below shows a solar panel, ABCD, which is fixed to a flat piece of concrete slab EFCD. ABCD and EFCD are two identical rhombuses. K is a point on DC such that DK=KC and AK⊥DC. AF and KF are drawn. AD^C=CD^E=60∘ and AD=x units.
7.1
Determine AK in terms of x.
(2)
7.2
Write down the size of KC^F.
(1)
7.3
It is further given that AK^F, the angle between the solar panel and the concrete slab, is y. Determine the area of ΔAKF in terms of x and y.