Christmas Streak 78/88: Geometry Rush #8

Notice that the angle $\angle BDC$ is two times the angle $\angle DCB$ . Let's call the angle $\angle DCB$ as $x$ . We can use the Law of Sines in the triangle $\triangle BCD$ :

$\frac{DC}{\sin x}=\frac{BC}{\sin 2x}=\frac{11}{\sin x}=\frac{19}{\sin 2x}$

$11 \sin 2x = 19 \sin x$

By the relation $\sin 2x= 2\sin x \cos x$ :

$11 \cdot 2\sin x \cdot \cos x = 19 \sin x$

$22\sin x \cos x = 19 \sin x$

$22\cos x = 19$

$\cos x = \frac{19}{22}$

Now we can use the Law of Cosines:

$11^2=19^2+BD^2-2\cdot19\cdot\frac{19}{22}$

$121=361+BD^2-\frac{722}{22}$

$BD^2-\frac{722}{22}+240$

Solving this equation we get $BD=11$ or $BD=\frac{240}{11}$ . The first solution is incorrect because it implies that $x=45$ and it isn't possible because there are no right triangle with legs measuring $11$ and hypotenuse measuring $19$ . The only possible solution is $BD=\frac{240}{11}$ .

Then you should input your answer as $240+11=251$