State the limits for B if given that B is an obtuse angle and

a) tan2B is negative

b) tan4B is positive

c) tan2B is negative or tan 4B is positive

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- Jun 19th 2013, 07:42 AMTrefoil2727limit for angle
State the limits for B if given that B is an obtuse angle and

a) tan2B is negative

b) tan4B is positive

c) tan2B is negative or tan 4B is positive - Jun 19th 2013, 08:19 AMPlatoRe: limit for angle
@Trefoil2727, Why do you keep posting a list of question, but show no effort?

From the given we know $\displaystyle \frac{\pi}{2}<B<\pi$. So any answer must be a subset of that.

a) if $\displaystyle \tan(2B)<0$ then it must be that $\displaystyle \frac{3\pi}{2}<2B<2\pi$

Now you do some work and show us. - Jun 20th 2013, 07:28 PMTrefoil2727Re: limit for angle
- Jun 21st 2013, 02:24 AMTrefoil2727Re: limit for angle
hah is that means that 360<4B<720?

so tan4B>0, 360<4B<(90+360), (180+360)<4B<(270+360)

90<B<112.5 , 135<4B<157.5 ?

tan2B is negative or tan 4B is positive means what? - Jun 21st 2013, 04:24 AMPlatoRe: limit for angle
It is clear to me that you don't understand much of this, do you?

From the given it must be true that $\displaystyle 90^o<B<180^o$. Thus any answer you find must be in that range.

If $\displaystyle \tan(2B)<0$ then $\displaystyle 270^o<2B<360^o$ or $\displaystyle 135^o<B<180^o$, which is in that range.

If $\displaystyle \tan(4B)>0$ then it could be $\displaystyle 180^o<4B<270^o$ or $\displaystyle 45^o<B<67.5^o$, which is**NOT in that range**.

So we must take another course.

Suppose that $\displaystyle 4B\in\text{quad } I$ or $\displaystyle 360^o<4B<450^o$ then $\displaystyle \tan(4B)>0$.

Solving we get $\displaystyle 90^o<B<112.5^o$, which is in the range.