Find $|BD|$ in the following quadrilateral...

Question

This is a preparation question for our national college entry exams;

Translating the paragraph;

$AB \perp BC$, $ACD$ is an equilateral triangle, $|AB|=2$, $|BC|=\sqrt{3}$,$|BD|=x$,,, Find $x$.

My Attempts

I have named $m(\widehat{CAB})=\alpha$, $\cos\alpha=\dfrac{2}{\sqrt7}$ and we can easily see that $|AD|=\sqrt{7}$ (1)

Then I've tried to calculate $x$ from cosines theorem in triangle $BAD$ (2)

I've calculated $\cos(60^\circ+\alpha)=\dfrac{\sqrt{3}}{2\sqrt{7}}$ (3)

Then I've applied the theorem: $4+7-2\cos(60^\circ+\alpha)=x^2$ (4)

What baffled me here a little is that we don't know whether $60+\alpha >90$ or not (5)

Let's pretend it isn^'t I said and then jumped to a conclusion like $x=\sqrt{11+\sqrt{3}}$ then I've tried to open it up a little more but I failed (9)

So the question is;

If my calculations are correct then this result isn't in the answers, is the question flawed? ( A little daring perhaps) If they are not how should I have done it, how could I know whether $60+\alpha>90$ or not? And the most important question, is there an easier way? (Perhaps one that skips all cosines and etc.) (I would be totally fine with a solution from area or sine theorem)

Thank you:))

Answer 1

by the Theorem of cosines we get $$BD^2=3+CD^2-6\cdot CD\cos(60^{\circ}+\alpha)$$(I) further $$CD=AC=\sqrt{7}$$ (II) $$\tan(\alpha)=\frac{2}{\sqrt{3}}$$ (III) Can you solve your Problem now?

Answer 2

You're probably making a calculation mistake. I've got x = 4 here. Your method is correct.

About the fact that whether cos(60 + alpha) is more than 0 or not, just expand it using the sum formula and plug in the values of sin alpha and cos alpha then you'll be able to figure out whether it is gonna be more than 0 or not.