How do we find the unknown angle using Law of Sines?
Given that
ABC is a triangle, $\hat{ABD} = 30^\circ, \hat{ACD} = \hat{DBC} = 10^\circ, \hat{DCB} = 20^\circ, \hat{BAD} = a$
Evaluate $a$
How do we solve this triangle using Law of Sines? I've seen something like
$$\biggr (\dfrac{\sin(110-a)}{\sin a}\biggr )\biggr (\dfrac{\sin80}{\sin10}\biggr )=1$$
but I dont know if that's related to Law of Sines.
$\endgroup$ 22 Answers
$\begingroup$It should be immediate that $\angle BAC = 110^\circ $. The Law of Sines applied to the three small triangles (the ones with $D$ as a vertex) gives you the following equalities:$$ \frac{\sin a }{\overline{BD}} = \frac{\sin 30}{\overline{AD}} ~~~~~~~~ \frac{\sin 20 }{\overline{BD}} = \frac{\sin 10}{\overline{CD}} ~~~~~~~~ \frac{\sin 10 }{\overline{AD}} = \frac{\sin (110 - a)}{\overline{CD}} $$It turns out that this is nice, because you can essentially cancel out the lengths of the sides. Rearrange the quotients as follows:$$ \frac{ \overline{AD}}{\overline{BD}} = \frac{\sin 30}{\sin a} ~~~~~~~~ \frac{\overline{BD}}{ \overline{CD}} = \frac{\sin 20}{\sin 10} ~~~~~~~~ \frac{\overline{CD}}{\overline{AD}} = \frac{\sin (110 - a)}{\sin 10 } $$If you multiply all the left sides together and multiply all the right sides together, you get $$ 1 = \frac{\sin 30 \sin 20 \sin (110 - a)}{\sin a \sin 10 \sin 10}$$From here, we'll be a little sneaky. First use $\sin 30 = \frac{1}{2}$ to get:$$ {\sin a \sin 10 \sin 10} = \frac{1}{2} \sin 20 \sin (110 - a)$$Now use the double-angle identity for $\sin 20$ and convert $\sin a$ to a cosine to get:$$ \cos (a - 90) \sin 10 = \cos 10 \sin(110 - a)$$From here, taking $\boxed{a = 100}$ solves the equation. Admittedly, this is a little hacky, but there may be a better way to get to the final answer after cancelling side lengths.
$\endgroup$ 7 $\begingroup$There is nothing to set the scale, so you can pick your favorite segment to be the unit. Say we define $BD=1$ Note $\angle BDC=150^\circ, \alpha+\angle DAC=110^\circ$. Then the Law of Sines says$$\frac {\sin \alpha}1=\frac {\sin 30}{AD}\\\frac {\sin (110-\alpha)}{CD}=\frac {\sin 10}{AD}$$You can keep going around the triangle until the sides divide out and you are left with a formula like you quote.
$\endgroup$
