Question

Sin a=2/5,Найти tg a.

Answer 1

если это вам не понятно =>| это корень

cos^2a=1-sin^2a

$cos \alpha = \sqrt{1 - \sin^{2} ( \alpha ) }$

cos a=| 1- 4/25=| 1-0,16= 0,84=

$\sqrt{ \frac{21}{25} } = \frac{ \sqrt{21} }{5}$

$\tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

$\tan( \alpha ) = \frac{ \frac{2}{5} }{ \frac{ \sqrt{21} }{5} } = \frac{2}{ \sqrt{21} }$

$\tan( \alpha ) = \frac{2}{ \sqrt{21} } = \frac{2 \sqrt{21} }{21}$

Answer 2

To find the tangent (tg) of angle "a," we can use the following formula:

tg(a) = sin(a) / cos(a)

Since we are given the value of sin(a) as 2/5, we need to find the value of cos(a) first. We can use the Pythagorean identity to do that:

sin^2(a) + cos^2(a) = 1

Now, substitute the given value of sin(a):

(2/5)^2 + cos^2(a) = 1

Solving for cos(a):

4/25 + cos^2(a) = 1 cos^2(a) = 1 - 4/25 cos^2(a) = 21/25

cos(a) = √(21/25) = √21 / 5

Now that we have the values of sin(a) and cos(a), we can find the tangent:

tg(a) = sin(a) / cos(a) tg(a) = (2/5) / (√21 / 5) tg(a) = (2/5) * (5/√21) tg(a) = 2/√21

To rationalize the denominator (remove the square root from the denominator), multiply both the numerator and denominator by √21:

tg(a) = (2/√21) * (√21/√21) tg(a) = 2√21 / 21

So, the value of tg(a) is 2√21 / 21.