Thread: Use discriminant to find the domain for which the angle is acute

I have the following equation -


and am required to use it for the problem - "Use the discriminant to find the domain for which the angle θ is acute." I'm having difficulty getting my head around the problem and understanding how the discriminant is useful to finding the solution. I'm sure it's something simple I'm just not seeing and I'll kick myself for not thinking of it when someone enlightens me. Any assistance would be greatly appreciated.

Thank you,
- deSitter

First, it should be obvious that $\displaystyle \displaystyle \begin{align*} x \neq 5 \end{align*}$. Now if angle $\displaystyle \displaystyle \begin{align*} \theta \end{align*}$ is acute, that means that $\displaystyle \displaystyle \begin{align*} \sin{(\theta)} > 0 \end{align*}$, giving

$\displaystyle \displaystyle \begin{align*} \frac{\sqrt{ (x - 10)(3x - 10)}}{2(5 - x)} &> 0 \\ \sqrt{( x - 10)(3x - 10)} &> 0 \\ (x - 10)(3x - 10) &> 0 \\ 3x^2 - 10x - 30x + 100 &> 0 \\ 3x^2 - 40x + 100 &> 0 \end{align*}$

If you want this equation to always be positive, that means it can never cross the x-axis, and so the equation $\displaystyle \displaystyle \begin{align*} 3x^2 - 40x + 100 = 0 \end{align*}$ can not have any solutions. What do you know about the discriminant when you don't have any solutions?

Hi Prove It, thank you for your reply. I'm a little hung up on "sin(θ) > 0", could this not also describe an Obtuse angle, as sin > 0 for [0, 180°]?

Actually I believe you are right, and also I forgot about the acute angles made in the fourth quadrant (for when $\displaystyle \displaystyle \begin{align*} -\frac{\pi}{2} \leq x \leq \frac{\pi}{2} \end{align*}$). But actually it doesn't affect my solution one bit. All that this tells us is that the sine value can not be 0 (or 1), and so that whole equation can't equal 0. The same rule for the discriminant applies.

Hello, deSitter!

$\displaystyle \sin\theta \:=\:\dfrac{\sqrt{(x-10)(3x-10)}}{2(5-x)}$

$\displaystyle \text{Use the discriminant to find the domain for which angle }\theta\text{ is acute.}$

$\displaystyle \theta$ is in Quadrant 1: .$\displaystyle 0 \,<\,\sin\theta \,<\,1$

We have: .$\displaystyle 0 \:<\:\dfrac{\sqrt{(x-10)(3x-10)}}{2(5-x)} \:<\:1$

We see that the numerator is always positive.
The denominator is positive when $\displaystyle \color{blue}{x \,<\,5}$


We have two inequalities to solve:


$\displaystyle [1]\;\;\dfrac{\sqrt{(x-10)(3x-10)}}{2(5-x)} \:<\:1 \quad\Rightarrow\quad \sqrt{(x-10)(3x-10)}} \:<\:2(5-x)$

. . $\displaystyle (x-10)(3x-10) \:<\:4(5-x)^2 \quad\Rightarrow\quad 3x^2 - 40x + 100 \:<\:100 - 40x + 4x^2$

. . $\displaystyle x^2 \:>\:0$

. . This is true for all $\displaystyle x \ne 0:\:\color{blue}{(\text{-}\infty,0) \cup (0,\infty)}$


$\displaystyle [2]\;\;\dfrac{(x-10)(3x-10)}{2(5-x)} \:>\:0 \quad\Rightarrow\quad (x-10)(3x-10) \:>\:0 $

. . $\displaystyle y \:=\:(x-10)(3x-10)$ is a parabola that opens upward.
. . It is positive (above the $\displaystyle x$-axis) "outside" of its $\displaystyle x$-intercepts.
. . . . $\displaystyle \color{blue}{x < \tfrac{10}{3}\,\cup\,x \,>\,10}$


$\displaystyle \text{Domain: }\:(\text{-}\infty,0)\cup \left(0,\tfrac{10}{3}\right)$

Got it! Thank you very much for your assistance.