How to find the angle $x$ using the law of sines?

Question

I'm trying to find the angle $x$ using the law of sines. The equalities I found are given below

$$\dfrac{\sin (16)}{|AD|} = \dfrac{\sin (x)}{|AB|} \tag{1}$$

$$ \dfrac{\sin (14)}{|DC|} = \dfrac{\sin (14)}{|DB|} = \dfrac{\sin (152)}{|BC|}\tag{2}$$

Could you assist me to take it from there?

Answer 1

Let $|BC|=a, |AB|=c, |AC|=b,$ and $|BD|=|CD|=d.$ In $\triangle BDC,$ we have$$ \begin{align} {\sin152^\circ\over a} &= {\sin14^\circ\over d}\\ {\sin(180^\circ-28^\circ)\over a} &= {\sin14^\circ\over d}\\ {\sin28^\circ\over a} &= {\sin14^\circ\over d}\\ {2\sin14^\circ\cos14^\circ\over a} &= {\sin14^\circ\over d}\\ a&=2d\cos14^\circ\tag{1} \end{align}$$

We see that $\angle A = 104^\circ,$ so in $\triangle ABC,$ we have $$ \begin{align} {\sin104^\circ\over a}&={\sin30^\circ\over b}\\ {\sin(90^\circ+14^\circ)\over a} &= {1\over2b}\\ {\cos14^\circ\over a}&={1\over2b}\\ a&=2b\cos14^\circ\tag{2} \end{align}$$ Comparing $(1)$ and $(2),$ we have $b=d.$

That means that $\triangle ADC$ is isosceles, and it's easy to work out the angles around point $D.$ I get $x=134^\circ$ (on the second try, thanks to Lance.)

Answer 2

OK. Sine law only approach.

WLOG, let $|AB|=1$.

$\sin x=\cfrac {\sin 16^\circ}{|AD|}$

$\cfrac{|AD|}{\sin 32^\circ}=\cfrac {|AC|}{\sin(360^\circ-x-152^\circ)}$

$|AC|=\cfrac {\sin 30^\circ}{\sin 46^\circ}$

From above, we have

$\sin x=\cfrac{\sin 16^\circ \sin(-x-152^\circ)\sin 46^\circ}{\sin 32^\circ \sin30^\circ}=\cfrac{2\sin 16^\circ \sin(x-28^\circ)\sin 46^\circ}{\sin 32^\circ }=\cfrac{\sin(x-28^\circ)\sin 46^\circ}{\cos 16^\circ } \tag 1$

expand $\sin 46^\circ$ doesn't seem to give any neat equation. Anyway, paul's solution is neat and insightful, except there is a minor arithmetic issue. Final result should be $x=134^\circ$.

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I got stuck here until I realized how to solve this from solving a similar problem.

$(1)\iff \cfrac{\sin x}{\sin (x-28^\circ)}=\cfrac {\sin 46^\circ}{\sin 74^\circ} \iff \cfrac{\sin x+\sin (x-28^\circ)}{\sin x-\sin (x-28^\circ)}=\cfrac {\sin 46^\circ+\sin 74^\circ}{\sin 46^\circ-\sin 74^\circ}$

$\iff \cfrac{2\sin(x-14^\circ)\cos 14^\circ}{2\cos(x-14^\circ)\sin 14^\circ}=\cfrac{2\sin 60^\circ\cos 14^\circ}{2\cos 60^\circ\sin(-14^\circ)}$

$\iff \tan (x-14^\circ)=-\tan 60^\circ=\tan 120^\circ$

$\iff x-14^\circ=120^\circ \iff x=134^\circ$

It's implicit that $0<x<180^\circ$

Answer 3

$$BDC=180-2(14)=152\tag{*},$$

$$\frac{\sin(30)}{AC}=\frac{\sin(46)}{AB}\tag{1},$$

$$\frac{\sin(x)}{AB}=\frac{\sin(16)}{AD}\tag{2},$$

and by (*),

$$\frac{\sin(208-x)}{AC}=\frac{\sin(32)}{AD}\tag{3}$$

$$\overset{\text{by }(2)}\implies\sin(x)=\sin(16)\frac{AB}{AD}\overset{\text{by }(1)}=\sin(16)\frac{\sin(46)}{\sin(30)}\frac{AC}{AD}\overset{\text{by }(3)}=\sin(16)\frac{\sin(46)}{\sin(30)}\frac{\sin(208-x)}{\sin(32)}.$$ Now apply $\sin(x-y)=\sin x\cos y-\cos x\sin y$ to obtain $$\sin(x)=\sin(16)\frac{\sin(46)}{\sin(30)}\frac{\cos(x)\sin(208)-\cos(208)\sin(x)}{\sin(32)}.$$ $$\sin(x)\left(1+\frac{\sin(16)\sin(46)\cos(208)}{\sin(30)\sin(32)}\right)=\frac{\sin(16)\sin(46)\sin(208)}{\sin(30)\sin(32)}\cos(x).$$ Hence, $$\tan(x)=\frac{\sin(16)\sin(46)\sin(208)}{\sin(30)\sin(32)+\sin(16)\sin(46)\cos(208)},$$ which gives $x=-46\implies x=-46+180=134$.

• All arguments in degrees.