Question Number 12713 by ashok kumar last updated on 29/Apr/17 | ||
$$\boldsymbol{{Q}}.\:\boldsymbol{\theta}\:=\:\mathrm{tan}^{−\mathrm{1}} \:\:\mathrm{4}/\mathrm{3}\: \\ $$$$ \\ $$ | ||
Commented by prakash jain last updated on 01/May/17 | ||
$$\mathrm{agar}\:\mathrm{ye}\:\mathrm{textbook}\:\mathrm{ka}\:\mathrm{question}\:\mathrm{hai} \\ $$$$\mathrm{to}\:\mathrm{trigonometric}\:\mathrm{tables}\:\mathrm{se}\:\mathrm{amswer} \\ $$$$\mathrm{aa}\:\mathrm{sakta}\:\mathrm{hai}. \\ $$ | ||
Answered by mrW1 last updated on 29/Apr/17 | ||
$$\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{4}/\mathrm{3}\right)\approx\mathrm{53}.\mathrm{13}° \\ $$ | ||
Commented by mrW1 last updated on 29/Apr/17 | ||
$${I}\:{don}'{t}\:{understand}\:{you}. \\ $$ | ||
Commented by ashok kumar last updated on 29/Apr/17 | ||
$${how}\:{to}\:{you}\:\mathrm{53}.\mathrm{13}° \\ $$ | ||
Commented by mrW1 last updated on 29/Apr/17 | ||
$${I}\:{used}\:{a}\:{calculator}\:{with}\:{trigonometric} \\ $$$${functions}. \\ $$$$ \\ $$$$\mathrm{tan}\:\mathrm{45}°=\mathrm{1} \\ $$$$\mathrm{tan}\:\mathrm{60}°=\sqrt{\mathrm{3}}\approx\mathrm{1}.\mathrm{732} \\ $$$${since}\:\mathrm{tan}\:\theta=\frac{\mathrm{4}}{\mathrm{3}}\approx\mathrm{1}.\mathrm{333} \\ $$$$\Rightarrow\theta\:{must}\:{be}\:{between}\:\mathrm{45}°\:{and}\:\mathrm{60}°. \\ $$ | ||
Commented by ashok kumar last updated on 29/Apr/17 | ||
$${kese}\:{aye}\:{sir}\:{please}\:{bato} \\ $$ | ||
Commented by prakash jain last updated on 01/May/17 | ||
$$\mathrm{Ashok},\:\mathrm{trigonometric}\:\mathrm{tables}\:\mathrm{se} \\ $$$$\mathrm{bhi}\:\mathrm{aa}\:\mathrm{sakta}.\:\mathrm{Kya}\:\mathrm{aapko}\:\mathrm{trigonometric} \\ $$$$\mathrm{tables}\:\mathrm{padna}\:\mathrm{aata}\:\mathrm{hai}? \\ $$ | ||