If $tan B = 3$ $tanB=3$ what is $sin B$ $sinB$ ?

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Answer 1 · 2016-12-15T22:29:33

Assuming you want your answer without the use of a calculator, $sin B = 3 cos B$ $sinB=3cosB$

$tan B = 3$ $tanB=3$

$\frac{sin B}{cos B} = 3$ $sinB/cosB=3$

$sin B = 3 cos B$ $sinB=3cosB$

But $cos B = \frac{1}{\sqrt{1 + {tan}^{2} B}} = \frac{1}{\sqrt{10}}$ $cosB=1/(sqrt(1+tan^2B))=1/sqrt10$

$sin B = \frac{3}{\sqrt{10}} = \frac{3}{10} \sqrt{10}$ $sinB=3/sqrt10=3/10sqrt10$

Answer 2 · 2016-12-15T22:51:17

We know that $tan θ = \frac{side opposite θ}{side adjacent θ}$ $tantheta = ("side opposite "theta)/("side adjacent "theta)$

So, the side opposite $B$ $B$ measures $3$ $3$ and the side adjacent $B$ $B$ measures $1$ $1$ .

We now find the length of the hypotenuse of the imaginary triangle:

$3^{2} + 1^{2} = h^{2}$ $3^2 + 1^2 = h^2$

$9 + 1 = h^{2}$ $9 + 1 = h^2$

$h = \sqrt{10}$ $h = sqrt10$

We now apply the definition that $sin θ = \frac{side opposite θ}{hypotenuse}$ $sintheta = ("side opposite "theta)/("hypotenuse")$ .

Therefore, $sin θ = \frac{3}{\sqrt{10}} = \frac{3 \sqrt{10}}{10}$ $sintheta = 3/sqrt(10) = (3sqrt(10))/10$ .

Hopefully this helps!

If tanB=3tanB=3 what is sinBsinB?