A triangle $ABC$ is such that $AC=\sqrt{2}$, $BC=4$ and angle $ACB=\frac{1}{4}\pi + heta $. Given t

Question

A triangle $ABC$ is such that $AC=\sqrt{2}$, $BC=4$ and angle $ACB=\frac{1}{4}\pi +	heta $. Given that $	heta $ is sufficiently small for ${{	heta }^{3}}$ and higher powers of $	heta $ to be neglected, show that $AB\approx \sqrt{10}\left[ 1+a	heta +b{{	heta }^{2}} ight]$, where $a$ and $b$ are real constants. [5]

Accepted Answer

$a=\frac25,\ b=\frac3{25}$

2021 DHS P1 Q2

2021 DHS P1 Q2

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