Question Number 31143 by NECx last updated on 03/Mar/18
$${A}\:{cyclist}\:{is}\:{travelling}\:{down}\:{a} \\ $$$${hill}\:{at}\:{a}\:{speed}\:{of}\:\mathrm{9}.\mathrm{2}{m}/{s}\:.\:{The} \\ $$$${hillside}\:{makes}\:{an}\:{angle}\:{of}\:\mathrm{6}.\mathrm{3}° \\ $$$${with}\:{the}\:{horizontal}\:.{Calculate}, \\ $$$${for}\:{the}\:{cyclist}: \\ $$$$\left({i}\right){the}\:{vertical}\:{speed} \\ $$$$\left({ii}\right){horizontal}\:{speed} \\ $$
Answered by MJS last updated on 03/Mar/18
$$\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{a}\:\mathrm{rectangular}\:\mathrm{triangle} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{9}.\mathrm{2}^{\mathrm{2}} \\ $$$$\alpha=\mathrm{6}.\mathrm{3}° \\ $$$$\beta=\left(\mathrm{90}−\mathrm{6}.\mathrm{3}\right)°=\mathrm{83}.\mathrm{7}° \\ $$$$\gamma=\mathrm{90}° \\ $$$$\frac{{a}}{\mathrm{sin}\:\alpha}=\frac{{b}}{\mathrm{sin}\:\beta}= \\ $$$${a}={c}×\mathrm{sin}\:\alpha \\ $$$${b}={c}×\mathrm{sin}\:\beta={c}×\mathrm{cos}\:\alpha \\ $$$${a}=\mathrm{9}.\mathrm{2sin}\:\mathrm{6}.\mathrm{3}=\mathrm{1}.\mathrm{009}\:\mathrm{555}…\:\mathrm{vertical}\:\mathrm{speed} \\ $$$${b}=\mathrm{9}.\mathrm{2cos}\:\mathrm{6}.\mathrm{3}=\mathrm{9}.\mathrm{144}\:\mathrm{440}…\:\mathrm{horizontal}\:\mathrm{speed} \\ $$
Commented by NECx last updated on 03/Mar/18
$${Thank}\:{you}\:{so}\:{much}. \\ $$