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2 years ago

Please help me whats -2(a-3)=12 showing your work and how you did it (solving for a)

Mathematics

1 answer:

kenny6666 [7]2 years ago

4 0

Answer:

i believed its a=-3

Step-by-step explanation:

-2(a-3)=12

-2a+6=12

-2a=6

divide -2

a=-3

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kramer

Using trigonometric identities, it is found that the sine and the tangent of the angle are given as follows:

$\sin{\theta} = \frac{1}{2}$ .

$\tan{\theta} = -\frac{\sqrt{3}}{3}$

<h3>How do we find the sine of an angle given the cosine?</h3>

We use the following identity:

$\sin^{2}{\theta} + \cos^{2}{\theta} = 1$

In this problem, the cosine is:

$\cos{\theta} = -\frac{\sqrt{3}}{2}$

Hence the sine is found as follows:

$\sin^{2}{\theta} + \left(-\frac{\sqrt{3}}{2}\right)^2 = 1$

$\sin^{2}{\theta} + \frac{3}{4} = 1$

$\sin^{2}{\theta} = \frac{1}{4}$

$\sin{\theta} = \pm \sqrt{\frac{1}{4}}$

Second quadrant, so the sine is positive, hence:

$\sin{\theta} = \frac{1}{2}$

<h3>What is the tangent of an angle?</h3>

The tangent is given by the <u>sine divided by the cosine</u>, hence:

$\tan{\theta} = \frac{\sin{\theta}}{\cos{\theta}}$

Hence:

$\tan{\theta} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}}$

$\tan{\theta} = -\frac{1}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$\tan{\theta} = -\frac{\sqrt{3}}{3}$

More can be learned about trigonometric identities at brainly.com/question/7331447

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