Question Number 90178 by I want to learn more last updated on 21/Apr/20
Commented by I want to learn more last updated on 21/Apr/20
$$\mathrm{I}\:\mathrm{got}\:\:\mathrm{26}.\mathrm{6}\:\mathrm{degree},\:\mathrm{please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{check}. \\ $$
Answered by TANMAY PANACEA. last updated on 21/Apr/20
$${tan}\mathrm{30}^{{o}} =\frac{\mathrm{100}−\mathrm{50}}{{base}} \\ $$$${base}=\mathrm{50}\sqrt{\mathrm{3}}\: \\ $$$${tan}\theta=\frac{\mathrm{100}}{\mathrm{50}\sqrt{\mathrm{3}}}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$${tan}\theta\approx{tan}\mathrm{49}^{{o}} \\ $$$$\theta\approx\mathrm{49}^{{o}} \\ $$
Commented by I want to learn more last updated on 21/Apr/20
$$\mathrm{Sir},\:\:\mathrm{which}\:\mathrm{angle}\:\mathrm{should}\:\mathrm{be}\:\mathrm{bigger} \\ $$
Commented by I want to learn more last updated on 21/Apr/20
$$\mathrm{Please}\:\mathrm{sir},\:\mathrm{the}\:\mathrm{diagram}. \\ $$
Commented by TANMAY PANACEA. last updated on 21/Apr/20
Commented by TANMAY PANACEA. last updated on 21/Apr/20
$${pls}\:{recheck} \\ $$
Commented by I want to learn more last updated on 21/Apr/20
$$\mathrm{I}\:\mathrm{draw}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{in}\:\mathrm{a}\:\mathrm{wrong}\:\mathrm{way}\:\mathrm{then}. \\ $$
Commented by I want to learn more last updated on 21/Apr/20
$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{appreciate}. \\ $$