Find the value of beta in degrees

[Richie Smash]

Homework Statement

In triangle ABC

AC= 9cm, BC=5cm and angle ACB = β in degrees.
Given that cos²β=0.84, determine
(i) the exact value of sin²B
(ii) the value of β if 90°<β<180°
(iii) the length of AB

Homework Equations

sin²θ+cos²θ=1

The Attempt at a Solution

Using that equation In relevant equations i calculated part (i) and it is 0.16

Now I'm stuck on part (ii), I tried finding the square root of sin²β and as well as cos²β
then use the inverse function.

But to no avail, what significance does this range play?

 

[LCKurtz]

Homework Statement

In triangle ABC

AC= 9cm, BC=5cm and angle ACB = β in degrees.
Given that cos²β=0.84, determine
(i) the exact value of sin²B
(ii) the value of β if 90°<β<180°
(iii) the length of AB

Homework Equations

sin²θ+cos²θ=1

The Attempt at a Solution

Using that equation In relevant equations i calculated part (i) and it is 0.16

Now I'm stuck on part (ii), I tried finding the square root of sin²β and as well as cos²β
then use the inverse function.

But to no avail, what significance does this range play?

You have $\cos^2\beta = .84$ so $\cos\beta = \pm\sqrt{.84} \approx \pm.92$. Which sign would work for $\beta$ in the second quadrant?

 

[SammyS]

 

[Richie Smash]

You have $\cos^2\beta = .84$ so $\cos\beta = \pm\sqrt{.84} \approx \pm.92$. Which sign would work for $\beta$ in the second quadrant?

Oh wow it was a simple matter of understanding the quadrants!
The range gives it away, it's clearly in the second quadrant and since cosine is the value its going to be the negative sign!
I got 156.4 degrees which is the answer, now time to work out the last part

 

[Richie Smash]

Hi this is a related question, but I drew a graph for a sin and tan function from 0 degrees to 60, and the value at which sinx=tanx is 0°

Now they ask derive the value for which sinx=tanx in the range 90°≤x≤270°

I can do this by memorization because yes of course I know sin 180° and tan 180°= 0°.

But I'm pretty sure that's not the method.

 

[LCKurtz]

Hi this is a related question, but I drew a graph for a sin and tan function from 0 degrees to 60, and the value at which sinx=tanx is 0°

Now they ask derive the value for which sinx=tanx in the range 90°≤x≤270°

I can do this by memorization because yes of course I know sin 180° and tan 180°= 0°.

But I'm pretty sure that's not the method.

I would do the algebra. Rewrite $\sin x = \tan x = \frac {\sin x}{\cos x}$ as $\sin x(\cos x -1)=0$ and ask yourself where on the given interval can either factor be $0$.

 

[Richie Smash]

Hmm, I'm not quite sure I'm following to be honest, How are you getting sinx(cosx-1)?
This has always been a big problem of mine, I have dificulty figuring out these range questions

 

[LCKurtz]

That's not a range problem, it is an algebra problem. Multiply both sides of $\sin x = \frac {\sin x}{\cos x}$ by $\cos x$ and go from there.

 

[Richie Smash]

Ooo now I understand how you got that expression.

Well it would be 180, but I just know what from knowledge, because sin 180 = 0

and sin180/cos180 is 0 / -1 = 0

 

[LCKurtz]

And you should also note that $\cos x = 1$ also can happen, and would solve the equation. Such values turn out to not be in your interval, but you still need to check them.