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This is from a grade-8 maths textbook:

In a quadrilateral $ABCD$, $AB=3.2 \text{cm}$, $BC=5.1 \text{cm}$, $\angle CBD=34.4°$, and the length of the diagonal $BD$ is $7.5 \text{cm}$. Given further that the area of $\triangle ABC$ is $11.62 \text{cm}^2$ and $\angle ABD$ is obtuse, find:

  1. the area of $\triangle BCD$,
  2. $\angle ABD$

Answer key says answer to

  • 1 is $10.8 \text{cm}^2$
  • 2 is $104.5°$
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  • $\begingroup$ So, it follows that $\sin \angle ABC=2*11.62/(3.2*5.1)\approx 1.424>1$? $\endgroup$ – g.kov Sep 20 '17 at 16:08
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for the first we get $$A_{BCD}=\frac{1}{2}\cdot 7.5cm\cdot 5.1cm\sin(34.4^{\circ})$$ for the second we get $$11.62cm^2=\frac{1}{2}\cdot 3.2cm\cdot 5.1cm\sin(\beta+34.4^{\circ})$$

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