the measure of one acute angle is twice the difference of the measure of the other acute angle and 12

1 answer:

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Is there a picture to find the measure of the acute angle? Or do you want an equation to solve it? If so than it is: 2b-12=a, where b is the measure of the other acute angle and a is the measure of the acute angle where it is twice the difference of the measurment of the other acute angle and 12.

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Step-by-step explanation:

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EXPLAIN THE SECOND PART.......

alexdok [17]

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$=\sin^2\theta\left(\dfrac1{\cos^2\theta}-1\right)$
$=\sin^2\theta(\sec^2\theta-1)$
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Now if $0^\circ$ , then both $\sin\theta>0$ and $\tan\theta>0$ . Adding two positive numbers gives another positive number, so $\tan\theta+\sin\theta>0$ . Why this is useful will be apparent shortly.

Back to the identity:

$\tan^2\theta-\sin^2\theta=\tan^2\theta\sin^2\theta$

Factorizing the left hand side, we have

$(\tan\theta-\sin\theta)(\tan\theta+\sin\theta)=\tan^2\theta\sin^2\theta$

Now, any number squared will be positive, which means the right hand side is necessarily greater than 0.

We showed earlier that $\tan\theta+\sin\theta>0$ . So we understand that we have

$\underbrace{(\tan\theta-\sin\theta)}_?\underbrace{(\tan\theta+\sin\theta)}_+=\underbrace{\tan^2\theta\sin^2\theta}_+$

The only way to multiply a number by a positive number to get yet another positive number is to have the first number also be positive, which means

$\tan\theta-\sin\theta>0$

and from this it follows that

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in the provided region.

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