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Given that

ABC is a triangle, $\hat{ABD} = 30^\circ, \hat{ACD} = \hat{DBC} = 10^\circ, \hat{DCB} = 20^\circ, \hat{BAD} = a$

Evaluate $a$

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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.

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  • $\begingroup$ The Law of Sines relates the sine of an angle and the side opposite as follows: $\frac{\sin A}{a} = \frac{\sin B}{b}$. $\endgroup$ – Sean Roberson Nov 27 '18 at 22:53
  • $\begingroup$ what does "solve the triangle" mean? determine all angles? $\endgroup$ – gt6989b Nov 27 '18 at 23:02
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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.

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  • $\begingroup$ Why and how did you decide the ratios of lenght is $1$? And can this be applied on other triangles like this? I also noticed that sometimes it is sufficient to know angles to apply Law of Sines. $\endgroup$ – Enzo Nov 28 '18 at 5:26
  • $\begingroup$ Multiplying all of the left-hand-sides gives $\frac{\overline{AD} \cdot \overline{BD} \cdot \overline{CD}}{\overline{AD} \cdot \overline{BD} \cdot \overline{CD}}$, which cancels out to become 1. The first part of the solution (up till this multiplication) can be applied to similar triangles, but the rest seems to work out by virtue of the specific values. $\endgroup$ – platty Nov 28 '18 at 5:30
  • $\begingroup$ Can't we figure it out without knowing the exact values? (Speaking for the rest) For instance, we can try to see what we can substitue in $a$ to equate the both sides. $\endgroup$ – Enzo Nov 28 '18 at 5:33
  • $\begingroup$ You can try, but it generally won't come out to something nice. But if you're looking for an approximate answer, you could always just plot the two functions and see where they intersect. $\endgroup$ – platty Nov 28 '18 at 5:43
  • $\begingroup$ the part where you claim $$ 1 = \frac{\sin 30 \sin 20 \sin (110 - a)}{\sin a \sin 10 \sin 10}$$ is just the result of Ceva's trigonometric form: artofproblemsolving.com/wiki/index.php?title=Ceva%27s_Theorem $\endgroup$ – Dr. Mathva Nov 28 '18 at 6:15
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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.

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