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Question Number 28583 by tawa tawa last updated on 27/Jan/18

Commented by tawa tawa last updated on 27/Jan/18

$$\mathrm{In}\mathrm{the}\mathrm{figure}\mathrm{above}:\mathrm{AB}\mathrm{is}\mathrm{a}\mathrm{chord}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{of}\mathrm{radius}10\mathrm{cm}.\mathrm{M}\mathrm{is}\mathrm{the}$$$$\mathrm{mid}\mathrm{point}\mathrm{of}\mathrm{AB}.\mathrm{and}\mathrm{NM}\mathrm{\perp }\mathrm{AB}$$

Answered by Rasheed.Sindhi last updated on 27/Jan/18

$$\mathrm{Let}\mathrm{O}\mathrm{is}\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{the}\mathrm{arc}.$$$$△\mathrm{OMA}\mathrm{is}\mathrm{right}\mathrm{angled}\mathrm{triangle}\mathrm{with}$$$$\mathrm{\angle }\mathrm{OMA}=90,\mathrm{OA}=\mathrm{r}\mathrm{cm},\mathrm{OM}=\mathrm{r}-2\mathrm{cm}$$$$\mathrm{AM}=\mathrm{x}\left(\mathrm{say}\right):$$$${(\mathrm{r}-2)}^2+{\mathrm{x}}^2={\mathrm{r}}^2$$$$\mathrm{x}=\sqrt{{\mathrm{r}}^2-{(\mathrm{r}-2)}^2}$$$$=\sqrt{4\mathrm{r}-4}=2\sqrt{\mathrm{r}-1}$$$$\mathrm{a}.\mathrm{AB}=2\mathrm{AM}=2\mathrm{x}=4\sqrt{\mathrm{r}-1}$$$$=4\sqrt{10-1}=12\mathrm{cm}$$$$\mathrm{b}.\mathrm{\angle }\mathrm{AOM}=\theta \left(\mathrm{Assumed}\right),\mathrm{OM}=10-2=8$$$$\mathrm{OA}=10$$$$\cos \theta =\frac{\mathrm{OM}}{\mathrm{OA}}=\frac{8}{10}=0.8$$$$\theta ={\cos }^{-1}(0.8)$$$$\mathrm{\angle }\mathrm{AOB}=2\mathrm{\angle }\mathrm{AOM}=2\theta =2{\cos }^{-1}(0.8)$$$$\approx 1.287\mathrm{rad}.$$$$\mathrm{c}.\mathrm{Let}\stackrel{⌢}{\mathrm{ANB}}=\mathrm{s}$$$$\mathrm{s}=\mathrm{r}\left(2\theta \right)=10\times 1.287=12.87$$$$\mathrm{Difference}=12.87-12=0.870\mathrm{cm}\approx 9\mathrm{mm}$$

Commented by tawa tawa last updated on 27/Jan/18

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}$$

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