How do we solve this triangle using law of sines?

Question

ABC is a right triangle, $AB \perp AC, DE \perp BC, |AD| = |BD| = 3$. The area of ABC is $6$ times of the area of BDE. Evaluate $|AC|$.

How do we solve this triangle using law of sines?

Answer 1

Let $\angle ABC = \theta$, then $EB = 3\cos\theta$ and $AB=6$. Since the two triangles are similar, the area scale factor is

$$ \frac{A_{BDE}}{A_{ABC}} = \left(\frac{EB}{AB}\right)^2 = \left(\frac{\cos\theta}{2}\right)^2 = \frac16 $$

$$ \implies \cos^2\theta = \frac23, \sin^2\theta = \frac13, \tan^2\theta = \frac12 $$

Finally

$$ AC = AB\tan\theta = \frac{6}{\sqrt{2}} = 3\sqrt{2} $$

Answer 2

We have $$A_{ABC}=6A_{BDE}$$ so we get $$6\frac{AC}{2}=6\frac{DE\cdot BE}{2}$$ with $$\frac{DE}{3}=\frac{AC}{BC}$$ and $$\frac{BE}{3}=\frac{6}{BC}$$ we get $$3AC=3\cdot 3\frac{AC}{BC}\cdot \frac{18}{BC}$$ so $$BC^2=54$$ and $$6^2+AC^2=54$$

Answer 3

$\triangle ABC$ and $\triangle DEB$ are similar.$m\angle ACB=m\angle EDB$. As per sine law,$\frac{c}{sinC}=\frac{a}{sinA}=\frac{b}{sinB}$ in $\triangle ABC$. Likewise, $\frac{d}{sinD}=\frac{b_1}{sinB}=\frac{e}{sinE}$in $\triangle DEB$.

For the sake of convenience, $\overline{AB}=c, \overline{AC}=b,\overline{BC}=a$ in $\triangle ABC$ and $\overline{DB}=e,\overline{BE}=d,\overline{DE}=b_1$ in $\triangle DEB$.

$m\angle E$ and $m\angle A$ are $90^\circ$ angle in $\triangle DEB$ and $\triangle ABC$ respectively.So, using sine law,

$\frac{c}{sinC}=\frac{b}{sinB}=\frac{a}{sinA}=a$ because $\sin{90^\circ}=1$.

Likewise, $\frac{d}{sinD}=\frac{b_1}{sinB}=\frac{3}{sinE}=3$ because $\sin{E}=\sin{90^\circ}=1$

$\sin{D}=\sin{C}$.

So, ${d}=3\sin{C}$ but $\sin{C}=\frac6a$.

So $d=\frac{18}{a}$. But area of $\triangle ABC$ is $6\times$ the area of $\triangle DEB$. It can be rewritten as $3b=3db_1$

So,$b=\frac{18}{a}b_1$ $\Rightarrow a*\sin{B}=\frac{54}{a}\sin{B}$

$a^2=54 \Rightarrow a=\sqrt{54}=7.348469$

$a^2-c^2=b^2$ So,$54-36=18,\Rightarrow b=\sqrt{18}$

Answer 4

A triangle is solved if there is sufficient information to apply a congruence rule, e.g. SSS, SAS, AAS, and HL.

We have that the ratio of areas between the larger and smaller triangles is 6 to 1. This implies the ratio if corresponding legs is $\sqrt{6}$:1.

We know the hypotenuse if the smaller triangle us 3. By the length ratio, the hypotenuse of the larger triangle us 3$\sqrt{6}$. Now the pythagorean theorem can tell us the length of the remaining unknown side.

By the sine rule, the sine of either angle divided by the opposite side is the reciprocal if the hypotenuse length. We know each leg length, so we can determine the sine, and by inverting, the angle itself