Trigonometry - Obtuse angle

Question

An obtuse angle $\theta$ has a $\sin \theta = \frac{x+1}{x+2}$ and $\cos \theta= \frac{x}{x+2}$. Which angle is it?

I've been frustrated with this for the last few hours now. My problem is that I can't seem to find the length of any of the sides. I've tried working it out with the $\sin \theta$ and $\cos \theta$ sentence, but I always get an answer which still contains something unknown - and that doesn't help much. Is it even possible to find the exact amount of degrees here? Or is it implied that I should only tell if it's A,B or C (Which seems weird to me).

Answer 1

You could use the fundamental relation:

$$\sin^2 \theta+\cos^2 \theta=1\to \left(\frac{x+1}{x+2}\right)^2+\left(\frac{x}{x+2}\right)^2=1\\ (x+1)^2+x^2=(x+2)^2\to x^2-2x-3=0\to \text{ $x=-1$ or $x=3$}$$

$1)$ For $x=-1$ you get

$$\sin \theta=0 \text{ and } \cos \theta=-1$$

what give you

$$\theta=(2k+1)\pi$$

Now check what you get when $x=3$.

Can you finish?

Answer 2

$$\sin \alpha = \frac{x+1}{x+2}, \,\cos \alpha=\frac{x}{x+2},\, \alpha\in \left(\frac{\pi}{2}, \pi\right), \, x\neq -2$$

Using $\sin^2\alpha+\cos^2\alpha=1$: $$1=\left(\frac{x+1}{x+2}\right)^2+\left(\frac{x}{x+2}\right)^2$$ $$1=\frac{2x^2+2x+1}{x^2+4x+4}$$ $${2x^2+2x+1}=x^2+4x+4$$ $${x^2-2x-3}=0$$ $$(x+1)(x-3)=0$$ So $x\in \{-1, 3\}$

For $x=-1$:

$\cos \alpha=\frac{-1}{1}=-1$

So $\alpha = \pi+2k\pi$, $k\in \mathbb{Z}$, but then $\alpha \not \in \left(\frac{\pi}{2}, \pi\right)$. Thus $x\neq -1$

For $x=3$

$\sin \alpha=\frac{4}{5}$

Then,as we assume, that $\alpha$ is obtuse, we have $$\alpha = \pi - \arcsin\frac{4}{5}$$

Answer 3

Avoid squaring as it introduces extraneous root(s).

$$1-\sin\theta=\dfrac1{x+2}=\dfrac{1-\cos\theta}2$$

$$\iff2\sin\theta=1+\cos\theta$$

$$\iff4\sin\dfrac\theta2\cos\dfrac\theta2=2\cos^2\dfrac\theta2$$

$$\iff2\cos\dfrac\theta2\left(\cos\dfrac\theta2-2\sin\dfrac\theta2\right)=0$$

Can you take it from here?