A sector arc contributes a linear term · source reference

Printed question

The diagram shows a triangle OPQ with OP = 5 cm, OQ = 7 cm and $\angle\mathrm{POQ}=\theta$ radians. Given that OP and OR are radii of a circle with centre O and $\theta$ is a sufficiently small angle, show that the perimeter of PQR can be approximated by $a+b\theta+c\theta^2$ where $a,b,c$ are constants to be determined. [5]

[2018/AJC/Prelim/I/2]

Printed working

Cosine rule: $\mathrm{PQ}^2=5^2+7^2-2(5)(7)\cos\theta$, hence $\mathrm{PQ}=\sqrt{74-70\cos\theta}$.

Perimeter $=\mathrm{PQ}+\mathrm{QR}+\text{arc PR}=\sqrt{74-70\cos\theta}+2+5\theta$.

$\approx[74-70(1-\theta^2/2)]^{1/2}+2+5\theta=(4+35\theta^2)^{1/2}+2+5\theta$.

$=2(1+35\theta^2/4)^{1/2}+2+5\theta\approx2(1+35\theta^2/8)+2+5\theta=4+5\theta+35\theta^2/4$.

$a=4,b=5,c=35/4$.

Teacher note 1

Use Cosine Rule. Factorise out 4 to make the constant term 1. Use $(AB)^n=A^nB^n$. Then use $(1+x)^{1/2}=1+x/2+\cdots$, replacing $x$ by $35\theta^2/4$.

Diagram 09-19-geometry

Arc PR is curved boundary radius 5 about O, not chord PR; R lies on OQ.

Mathematical review

$PQ^2=5^2+7^2-2(5)(7)\cos\theta=74-70\cos\theta$. Since $\cos\theta=1-\theta^2/2+O(\theta^4)$, $PQ=\sqrt{4+35\theta^2+O(\theta^4)}=2+35\theta^2/4+O(\theta^4)$.

Perimeter $=PQ+QR+\text{arc }PR=4+5\theta+35\theta^2/4+O(\theta^4)$ cm. Thus $a=4,b=5,c=35/4$. The curved boundary is an arc, not the straight chord joining P and R.

Provenance

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