# Trigonometry Question

ABCD is an isosceles trapezoid.

$$CAB = \alpha,\quad CAD = \beta,\quad AC = m$$

a) Find $AB$ and $DC$.
b) $\beta = 2\alpha$, $\frac{AB}{DC}$ = $\frac{1}{2}$, Find $a$ without using $β$ or $m$.
c) Is it possible to inscribe a circle in $ABCD$? Why?

a) $AB = \frac{m\sin(2\alpha+\beta)}{\sin(\alpha+\beta)}$, $DC = \frac{m\sin(\beta)}{\sin(\alpha+\beta)}$

b) $\alpha = 37.76^\circ$

c) No.

I've managed to solve a, but I'm having trouble solving b.

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Part (b) follows directly from part (a). We have $\dfrac{AB}{DC}=\dfrac{1}{2}$. Using the expressions you got for $AB$ and $CD$, putting $\beta=2\alpha$, and simplifying a bit, we find that $$\frac{1}{2}=\frac{AB}{CD}=\frac{\sin 4\alpha}{\sin 2\alpha}$$ (there is a fair amount of cancellation.)

Now use the double-angle identity $\sin 4\alpha=2(\sin 2\alpha)(\cos 2\alpha)$. We find that $$\cos 2\alpha =\frac{1}{4}.$$ The rest is a job for the calculator, use the $\cos^{-1}$ button.

If we feel like it we can get an explicit expression for $\cos \alpha$ or $\sin \alpha$, from the fact that $\cos 2\alpha=2\cos^2\alpha-1=1-2\sin^2\alpha$, but that is not necessary.

As to part (c), one cannot inscribe a circle in any trapezoid that satisfies the conditions of (b). First draw an isosceles trapezoid with an inscribed circle. Suppose that the two tangents to a circle from an external point $P$, meet the circle at $M$ and $N$. Then $PM=PN$. So in any isosceles trapezoid with an inscribed circle, each "slant" side has length equal to half the sum of the two parallel sides.

In our case, we can take $AB=2$ and $CD=4$. So if our trapezoid had an inscribed circle, then each slant side would have length $3$. But then the height of the trapezoid would be $\sqrt{3^2-1^2}$, that is, $2\sqrt{2}$. That makes $\tan\alpha =\frac{2\sqrt{2}}{3}$. But then $\angle \alpha\approx 43.3$ degrees, which is not what we found in part (b).

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Could you please briefly explain how you got to AB/CD = sin4a/sin2a? – Lior Apr 13 '12 at 18:49
From your work, putting $\beta=2\alpha$, we have $AB=\sin 4\alpha/(m\sin 3\alpha)$ and $CD=\sin 2\alpha/(m\sin 3\alpha)$. Divide $AB$ by $CD$. The denominators cancel, and we just end up with $\sin 4\alpha/\sin 2\alpha$. – André Nicolas Apr 13 '12 at 18:59
Thanks, obvious now, I don't know what I was thinking :) – Lior Apr 13 '12 at 19:03
Oh, and as for (c), why can't a circle be inscribed in the trapezoid? – Lior Apr 13 '12 at 19:08
@Lior: I added a solution of (c). Roughly speaking, there is only one isosceles trapezoid with top and bottom in ratio $1:2$ that has an inscribed circle. I then found what $\alpha$ must be, and it is not what we found in (b). One could also compute $\beta$ in this case and show it is not $2\alpha$, but one solution is enough! – André Nicolas Apr 13 '12 at 20:06