Question

Выразите tg70° через m, если m=12sin5°cos5°cos10°

Answer 1

m = 12Sin5°Cos5°cos10° = 6 * (2Sin5°Cos5°) * Cos10° = 6Sin10°Cos10° =

= 3 * (2Sin10°Cos10°) = 3Sin20°

Sin20° = 1/3m

tg70° = tg(90° - 20°) = Ctg20°

$1+Ctg^{2}20^{o}=\frac{1}{Sin^{2}20^{o}} \\\\Ctg^{2}20^{o} =\frac{1}{Sin^{2}20^{o}}-1=\frac{1}{(\frac{m}{3})^{2}}-1=\frac{9}{m^{2} }-1=\frac{9-m^{2}}{m^{2}}\\\\tg70^{o}=Ctg20^{o}=\frac{\sqrt{9-m^{2}}}{m}$

Answer 2

$m = 12 \times \sin5 \times \cos5 \times \cos10 = 6 \times (2 \sin5 \times \cos5) \times \cos10 = \\ \\ = 6 \sin10 \times \cos10 = 3 \times (2 \sin10 \times \cos10) = 3 \sin20 \\ \\ m = 3 \sin20 \: \: \: = \: \: \: sin20 = \frac{m}{3} \\ \\ tg70 = ctg(90 - 20) = ctg20 \\ \\ ctg20 = \frac{ \cos20 }{ \sin20 } = \frac{ \sqrt{1 - {( \sin20) }^{2} } }{ \sin20 } = \frac{ \sqrt{1 - { (\frac{m}{3} )}^{2} } }{ \frac{m}{3} } = \\ \\ = \frac{ \sqrt{ \frac{9 - {m}^{2} }{9} } }{ \frac{m}{3} } = \frac{ \sqrt{9 - {m}^{2} } }{3} \times \frac{3}{m} = \frac{ \sqrt{9 - {m}^{2} } }{m} \: \: \: \: \: (otvet) \\ \\$