Question 855017
If angles A and B are acute, altitude CD divides the triangle, into two right triangles:
{{{drawing(300,300,-5,3,-1.5,6.5,
green(rectangle(0,0,0.3,0.3)),
triangle(-4,0,2,0,0,5),
green(line(0,0,0,5)),
locate(-4.2,-0.1,A),locate(2,-0.1,B),
locate(-0.1,-0.1,D),locate(-0.1,5.5,C),
locate(-2,2.5,b),locate(0.8,2.5,a)
)}}} 
In those right triangles, the trigonometric ratios involving altitude CD and angles A and B are
{{{sin(A)=CD/b}}}<-->{{{CD=b*sin(A)}}}
{{{sin(B)=CD/a}}}<-->{{{CD=a*sin(B)}}}
 
If one of the angles A and B is obtuse, sides AC and BC, along with altitude CD, and the line containing AB form two right triangles:
{{{drawing(300,300,-5,3,-1.5,6.5,
green(rectangle(2,0,1.7,0.3)),
triangle(-4,0,0,0,2,5),
green(line(2,0,2,5)),green(arrow(0,0,3,0)),
locate(-4.2,-0.1,A),locate(2,-0.1,D),
locate(-0.1,-0.1,B),locate(1.9,5.5,C),
locate(-1,2.5,b),locate(0.6,2.5,a)
)}}} Ray BD is the extension of AB. Obtuse angle ABC is called angle B for short.
Angle B and angle CBD are supplementary and therefore have the same sine.
In right triangles ADC and BDC, the trigonometric ratios involving altitude CD and angles A and B are
{{{sin(A)=CD/b}}}<-->{{{CD=b*sin(A)}}}
{{{sin(B)=sin(CBD)=CD/a}}}<-->{{{CD=a*sin(B)}}}
 
In either case, {{{CD=b*sin(A)}}} and {{{CD=a*sin(B)}}} ,
so {{{b*sin(A)=a*sin(B)}}}
Dividing both sides of the equal sign by {{{ab}}} we get
{{{sin(a)/a=sin(B)/b}}}