Vrftsjtryk

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?

Let $|BC|=a, |AB|=c, |AC|=b,$ and $|BD|=|CD|=d.$ In $triangle BDC,$ we have$$
begin{align}
{sin152^circover a} &= {sin14^circover d}\
{sin(180^circ-28^circ)over a} &= {sin14^circover d}\
{sin28^circover a} &= {sin14^circover d}\
{2sin14^circcos14^circover a} &= {sin14^circover d}\
a&=2dcos14^circtag{1}
end{align}$$

We see that $angle A = 104^circ,$ so in $triangle ABC,$ we have $$
begin{align}
{sin104^circover a}&={sin30^circover b}\
{sin(90^circ+14^circ)over a} &= {1over2b}\
{cos14^circover a}&={1over2b}\
a&=2bcos14^circtag{2}
end{align}$$
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.)

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

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

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

and by (*),

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

$$overset{text{by }(2)}impliessin(x)=sin(16)frac{AB}{AD}overset{text{by }(1)}=sin(16)frac{sin(46)}{sin(30)}frac{AC}{AD}overset{text{by }(3)}=sin(16)frac{sin(46)}{sin(30)}frac{sin(208-x)}{sin(32)}.$$
Now apply $sin(x-y)=sin xcos y-cos xsin y$ 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 $x=-46implies x=-46+180=134$.

• All arguments in degrees.

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

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

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 $x=134^circ$.

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

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

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

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

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

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