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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

$$\mathrm{I}\mathrm{got}26.6\mathrm{degree},\mathrm{please}\mathrm{help}\mathrm{me}\mathrm{check}.$$

Answered by TANMAY PANACEA. last updated on 21/Apr/20

$$tan{30}^o=\frac{100-50}{base}$$$$base=50\sqrt{3}$$$$tan\theta =\frac{100}{50\sqrt{3}}=\frac{2}{\sqrt{3}}$$$$tan\theta \approx tan{49}^o$$$$\theta \approx {49}^o$$

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

$$plsrecheck$$

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

$$\mathrm{Thanks}\mathrm{sir},\mathrm{i}\mathrm{appreciate}.$$

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