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Problem: In the parallelogram $ABCD, \quad \measuredangle DBC=45^{\text{o}}, \quad\measuredangle BDC=30^{\text{o}}$ and $|AB|=5\sqrt{2}$. Find the length of $AD.$

I drew this:

enter image description here

Is this interpretation correct? I know that $\measuredangle BAD=\measuredangle BCD=105^{\text{o}},$ using the law of cosines I get

$$|BD|^2=x^2+(5\sqrt{2})^2-10x\sqrt{2}\cos{105^{\text{o}}}.$$

I also know that $\measuredangle ABD=30^{\text{o}},$ so now I can apply law of cosines again on $\triangle ABD$ with sides $BD, \quad 5\sqrt{2}$ and the angle between the two sides $30^{\text{o}}.$ So:

$$x^2=|BD|^2+(5\sqrt{2})^2-10|BD|\sqrt{2}\cos{30^{\text{o}}},$$

but this is a polynomial of the 4th degree with only complex solutions. I'm probably making it a lot harder than it is. Any tips?

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    $\begingroup$ Or you can apply the law of sines once. $\endgroup$ – Kenny Lau Aug 27 '17 at 10:12
  • $\begingroup$ @KennyLau - How? $\endgroup$ – Parseval Aug 27 '17 at 10:14
  • $\begingroup$ See my answer . $\endgroup$ – Kenny Lau Aug 27 '17 at 10:14
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$$\begin{array}{rcl} \dfrac{AD}{\sin \angle ABD} &=& \dfrac{AB}{\sin \angle ADB} \\ \dfrac{x}{\sin 30^\circ} &=& \dfrac{5\sqrt2}{\sin 45^\circ} \\ 2x &=& 10 \\ x &=& 5 \\ \end{array}$$

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  • $\begingroup$ Embarrasingly easy. Ofcourse, thank you! $\endgroup$ – Parseval Aug 27 '17 at 10:16
  • $\begingroup$ Sorry, I thought I already did. $\endgroup$ – Parseval Aug 27 '17 at 10:31
  • $\begingroup$ @Parseval thanks. $\endgroup$ – Kenny Lau Aug 27 '17 at 10:34
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Note that $AD=BC$ and $AB=CD$

$CD/\text{sin}45^\circ=x/\text{sin}30^\circ$

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Let $CE$ be an altitude of $\Delta DEC$.

Thus, since $\measuredangle EDC=30^{\circ}$, we obtain $$EC=\frac{1}{2}\cdot5\sqrt2=\frac{5}{\sqrt2}$$ and since $\measuredangle DBC=45^{\circ}$, we obtain: $$BC=EC\cdot\sqrt2=5.$$ Done!

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