Wiedząc, że tg alfa=2,4 i kąt alfa jest ostry, wyznacz wartość wyrażenia 2sin alfa+cos alfa/ 5sin alfa -2 cos alfa

[latex]tgalpha =2,4=frac{24}{10}=frac{12}{5} \frac{sinalpha }{cosalpha }=frac{12}{5} Rightarrow sinalpha =frac{12}{5}cosalpha \sin^2alpha +cos^2alpha =1 \frac{144}{25}cos^2alpha +cos^2alpha =1 \ frac{169}{25}cos^2 alpha =1 \cos^2 alpha = frac{25}{169} \cos alpha = frac{5}{13} \sin alpha = frac{12}{5} cdot frac{5}{13} = frac{12}{13} [/latex] [latex]\\ frac{2sin alpha +cos alpha }{5sin alpha -2cos alpha } = frac{ frac{24}{13}+ frac{5}{13} }{ frac{60}{13}- frac{10}{13} } = frac{29}{13}cdot frac{13}{50} = frac{29}{50}= frac{58}{100}=0,58 [/latex]