Question Number 150864 by puissant last updated on 16/Aug/21
Answered by ajfour last updated on 16/Aug/21
$$\mathrm{70}°=\theta,\:\mathrm{30}°=\alpha,\:\mathrm{40}°=\beta,\:\mathrm{6}{cm}={r} \\ $$$${A}_{\mathrm{1}} =\left({r}−\mathrm{2}{r}\mathrm{cos}\:\theta\right){r}\mathrm{sin}\:\beta \\ $$$${A}_{\mathrm{2}} =\frac{{r}\left(\mathrm{1}−\mathrm{2cos}\:\theta\right)\left\{{r}\mathrm{sin}\:\theta−{r}\mathrm{sin}\:\beta\right\}}{\mathrm{2}} \\ $$$${A}_{\mathrm{3}} =\frac{{r}^{\mathrm{2}} }{\mathrm{2}}\left(\frac{\pi}{\mathrm{6}}\right)−\frac{{r}}{\mathrm{2}}\left({r}\mathrm{sin}\:\alpha\right) \\ $$$${B}={A}_{\mathrm{1}} +{A}_{\mathrm{2}} +{A}_{\mathrm{3}} \\ $$
Commented by ajfour last updated on 16/Aug/21
$${give}\:{options}.. \\ $$