Wotzdorfg

2021-08-19

${\mathrm{tan}40}^{\circ }=\frac{x}{y}$
${\mathrm{tan}26}^{\circ }=\frac{x}{y+3.9}$
Find x

Mayme

${\mathrm{tan}40}^{\circ }=\frac{x}{y}\to y=\frac{x}{{\mathrm{tan}40}^{\circ }}$
${\mathrm{tan}26}^{\circ }=\frac{x}{y+3.9}\to y+3.9=\frac{x}{{\mathrm{tan}26}^{\circ }}$
Substitute to solve for x:
$\frac{x}{{\mathrm{tan}40}^{\circ }}+3.9=\frac{x}{{\mathrm{tan}26}^{\circ }}$
$\frac{x}{{\mathrm{tan}26}^{\circ }}-\frac{x}{{\mathrm{tan}40}^{\circ }}=3.9$
$x\left(\frac{1}{{\mathrm{tan}26}^{\circ }}-\frac{1}{{\mathrm{tan}40}^{\circ }}\right)=3.9$
$x=\frac{3.9}{\frac{1}{{\mathrm{tan}26}^{\circ }}-\frac{1}{{\mathrm{tan}40}^{\circ }}}$
$x\approx 4.5m$

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