Mathematics: The heights of the rose garden are 98 cm, 97 cm, 100 cm, 109 cm, 101 cm. what is the average height of the rose bushes in the garden

The heights of the rose garden are 98 cm, 97

alarconanais07

06.05.2022

$\tan(\theta)=\frac{-\sqrt{5}}{2}$

Step-by-step explanation:

Since we are in quadrant 2, sine is positive. Since sine is positive and cosine is negative, then tangent is negative.

Now I'm going to find the sine value of this angle given using one of the Pythagorean Identities, namely $\sin^2(\theta)+\cos^2(\theta)=1$ .

If given $\cos(\theta)=\frac{-2}{3}$ , then we have $\sin^2(\theta)+(\frac{-2}{3})^2=1$ by substitution of $\cos(\theta)=\frac{-2}{3}$ .

Let's solve:

$\sin^2(\theta)+(\frac{-2}{3})^2=1$ for $\sin(\theta)$ .

$\sin^2(\theta)+\frac{4}{9}=1$

Subtract 4/9 on both sides:

$\sin^2(\theta)=1-\frac{4}{9}$

Simplify:

$\sin^2(\theta)=\frac{5}{9}$

Square root both sides:

$\sin(\theta)=\sqrt{\frac{5}{9}}$

$\sin(\theta)=\frac{\sqrt{5}}{\sqrt{9}}$

$\sin(\theta)=\frac{\sqrt{5}}{3}$

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$\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}=\frac{\frac{\sqrt{5}}{3}}{\frac{-2}{3}}$

Multiplying top and bottom by 3 gives:

$\tan(\theta)=\frac{\sqrt{5}}{-2}$

I'm going to move the factor of -1 to the top:

$\tan(\theta)=\frac{-\sqrt{5}}{2}$

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