Question

90 POINTS PLEASE HELP

Answer 1

$sin\alpha = - 0.8$

Step-by-step explanation:

We have , $cos\alpha = \frac{3}{5}$ , where $\alpha$ is located in IV quadrant! Let's find out value of $sin\alpha$ :

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

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

⇒ $sin\alpha = \sqrt{1-(\frac{3}{5} )^2$

⇒ $sin\alpha = \sqrt{1-(\frac{9}{25} )$

⇒ $sin\alpha = \sqrt{(\frac{25-9}{25} )$

⇒ $sin\alpha = \sqrt{(\frac{16}{25} )$

⇒ $sin\alpha = \pm {(\frac{4}{5} )$

Value of $sin\alpha$ is dependent on which quadrant it is . Since, in question it's given that $\alpha$ is located in IV quadrant , So $sin\alpha$ is negative i.e.

⇒ $sin\alpha = - {(\frac{4}{5} )$

⇒ $sin\alpha = - 0.8$

Therefore, $sin\alpha = - 0.8$ .

Answer 2

Answer:

sin/alpha = -0.8

Step-by-step explanation: