Question

$\sf if \:tan \theta =-\dfrac{4}{3},then\:sin \theta \:is$

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Answer 1

Given:-

$\tan( \alpha ) = - \frac { 4}{3}$

To find:-

$\sin( \alpha ) =$ ?

Solution:-

Now perpendicular = -4 and base = 3

(Hypotenuse)² = (perpendicular)²+(base)²

(Hypotenuse)² = 3²+4² = 9 + 16 = 25

Hypotenuse = √25 = 5

Hypotenuse = 5

$\sin( \alpha ) = - \frac{4}{5}$

Hope its help uh❤

Answer 2

Given :-

$\sf{tan\theta=-\dfrac{4}{3}}$

To Find :-

The value of $\sf{sin\theta}$.

Solution :-

We know,

$\boxed{\bf{tan\theta=\frac{perpendicular}{base}}}$

$\displaystyle \sf{\longmapsto tan\theta=-\frac{4}{3} }$

∴ p = -4, b = 3

Using Pythagoras Theorem :-

h² = p² + b²

⇒ h = √[(-4)² + (3)²]

⇒ h = √(16 + 9)

⇒ h = √25

⇒ h = 5

We know,

$\boxed{\bf{sin\theta=\frac{perpendicular}{hypotenuse}}}$

$\displaystyle \sf{\longmapsto sin\theta=-\frac{4}{5} }$

Therefore, the answer is -4/5.