(CIE 0606/2020/w/22/Q8)
DO NOT USE A CALCULATOR IN THIS QUESTION.
In this question lengths are in centimeters.
(a) Given that the area of the triangle $ABC$ is 5.5 cm$^2,$ find the exact length of $AC.$ Write your answer in the form $a+b\sqrt 3$ , where $a$ and $b$ are integers. [4]
(b) Show that $BC^2= c+d\sqrt 3 $, where $c$ and $d$ are integers to be found. [4]
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*********math solution************* (a) $\begin{array}[t]{rcll} \dfrac{1}{2}(2 \sqrt{3}+1) A C \sin 30^{\circ}&=&5.5 \\(2 \sqrt{3}+1) A C&=&5.5\times 2\times \sin 30^{\circ}=22 \\A C&=&\dfrac{22}{(2 \sqrt{3}+1)} \times \dfrac{(2 \sqrt{3}-1)}{(2 \sqrt{3}-1)} \\A C&=&4 \sqrt{3}-2 \end{array}$ (b) $\begin{array}[t]{rcll}B C^{2}&=&(2 \sqrt{3}+1)^{2}+(4 \sqrt{3}-2) ^2-2(2 \sqrt{3}+1)(4 \sqrt{3}-2) \cos 30^{\circ}\\&=&(12+4\sqrt 3+1)+(48-16\sqrt 3+4)-2(24-4\sqrt 3+4\sqrt 3-2) \dfrac{\sqrt 3}{2}\\&=&[13+4 \sqrt{3}]+[52-16 \sqrt{3}] +[-22 \sqrt{3}]\\ &=&65-34 \sqrt{3} \end{array}$ **********end math solution******************** |
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