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Determine any point in the third quadrant for which tan(α)=4.

One valid point = ??

For this angle, sin(α) = ??

Thank you for your help!!

Respuesta :

Answer:

The point in third quadrant is: (-1,-4)

The value of sinα is [tex]\dfrac{-4}{\sqrt{17}}[/tex]

Step-by-step explanation:

We are given that:

[tex]\tan \alpha =4[/tex]

As we know that

[tex]\tan \alpha =\dfrac{P}{B}[/tex]

Where P denotes the perpendicular and B denote the base of a right angled triangle formed by the help of the trignometric identity.

Thus means that P=4 and B=1

Hence the hypotenuse is given by H=[tex]\sqrt{17}[/tex] ( with the help of Pythagorean theorem)

so we could construct a triangle in third quadrant

Hence the point in third quadrant is (-1,-4).

Now corresponding to this point [tex]\sin \alpha[/tex] is given as:

[tex]\sin \alpha=\dfrac{P}{H}[/tex]

⇒  [tex]\sin \alpha=\dfrac{-4}{\sqrt{17}}[/tex] (Negative sign is used as [tex]\sin[/tex] is negative in the third quadrant)



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Answer:

The valid point in third quadrant is  (-1, -4)

For this angle,  [tex]sin(\alpha)=\frac{-4}{\sqrt{17}}[/tex]

Step-by-step explanation:

Given that,   [tex]tan(\alpha)= 4[/tex]

We know that, [tex]tan=\frac{Opposite}{Adjacent}[/tex]

According to the below diagram, as [tex]\alpha[/tex] is in third quadrant, so....

[tex]tan(\alpha)= \frac{AB}{BO}=\frac{-4}{-1}[/tex]

That means,  [tex]AB=-4[/tex] and [tex]BO= -1[/tex]

So, the coordinate of point A will be:  [tex](-1, -4)[/tex]


Now,  [tex]Sin(\alpha)=\frac{Opposite}{Hypotenuse}=\frac{AB}{AO}[/tex]

Using Pythagorean theorem.......

[tex]AO^2= AB^2+BO^2\\ \\ AO^2= (-4)^2+(-1)^2\\ \\ AO^2= 16+1=17\\ \\ AO=\sqrt{17}[/tex]

So,  [tex]Sin(\alpha)= \frac{AB}{AO}= \frac{-4}{\sqrt{17}}[/tex]

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