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marysya [2.9K]
3 years ago
7

90 POINTS PLEASE HELP

Mathematics
2 answers:
11111nata11111 [884]3 years ago
4 0

sin\alpha  = - 0.8

<u>Step-by-step explanation:</u>

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 .

Arturiano [62]3 years ago
4 0

Answer:

sin/alpha = -0.8

Step-by-step explanation:

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