Question Number 46636 by azharkhan250963@gmail.com last updated on 29/Oct/18
$$\mathrm{tan}\:\theta=\mathrm{10tan60}^{°} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Oct/18
$${tan}\theta=\mathrm{10}×\sqrt{\mathrm{3}}\: \\ $$$${tan}\theta=\mathrm{10}×\mathrm{1}.\mathrm{732} \\ $$$$\theta={tan}^{−\mathrm{1}} \left(\mathrm{17}.\mathrm{32}\right) \\ $$