Question 827006
Question 827006
<pre>
AP, PQ and QR are three equal line segments inclined at angles
{{{alpha}}},{{{2alpha}}}, and{{{3alpha}}} respectively to line
segment AB.  Show that

tan(< BAR) = {{{RB/AB}}} = {{{(sin(alpha)+sin(2alpha)+sin(3alpha))/(cos(alpha)+cos(2alpha)+cos(3alpha))}}}


{{{drawing(400,2400/7,-.5,3,-.5,2.5,

line(0,0,2.2743,0),line(0,0,.9397,.342),line(.9397,.342,1.7057,.9848),
line(1.7057,.9848,2.2743,1.8508),green(line(.5321,0,.9397,.342),
line(1.2057,0,1.7057,.9848)), locate(0,0,A),locate(2.2743,0,B),
locate(.86,.5,P), locate(1.6,1.1,Q),locate(2.25,2,R),
locate(.34,.15,alpha),locate(.68,.15,2alpha), locate(1.3,.15,3alpha),
red(line(0,0,2.2743,1.8508),line(2.2743,1.8508,2.2743,0))


  )}}}

Draw PS and QT perpendicular to AB, PU and QV parallel to AB.
{{{drawing(400,2400/7,-.5,3,-.5,2.5,

line(0,0,2.2743,0),line(0,0,.9397,.342),line(.9397,.342,1.7057,.9848),
line(1.7057,.9848,2.2743,1.8508),green(line(.5321,0,.9397,.342),
line(1.2057,0,1.7057,.9848)), locate(0,0,A),locate(2.2743,0,B),
locate(.86,.5,P), locate(1.6,1.1,Q),locate(2.25,2,R),
locate(.34,.15,alpha),locate(.68,.15,2alpha), locate(1.3,.15,3alpha),
red(line(0,0,2.2743,1.8508),line(2.2743,1.8508,2.2743,0)),
blue(line(.9397,.342,.9397),line(1.7057,.9848,1.7057,0)),locate(.92,0,S),
locate(1.68,0,T),blue(line(.9397,.342,1.7057,.342),
line(1.7052,.9448,2.2743,.9448)),
locate(2.3,1,V),locate(1.74,.43,U)


  )}}}

{{{PS/AP=sin(alpha)}}}

{{{PS=AP*sin(alpha)}}}

< QPU = {{{2alpha}}}  Parallel lines cut by transversal

{{{QU/PQ=sin(QPU)=sin(2alpha)}}}

{{{QU=PQ*sin(2alpha)}}}

{{{QT=QU+UT=QU+PS=PQ*sin(2alpha)+AP*sin(alpha)}}}

{{{RV/QR=sin(3alpha)}}}

{{{RV=QR*sin(3alpha)}}}

{{{RB = RV+VB = RV+QT = QR*sin(3alpha)+PQ*sin(2alpha)+AP*sin(alpha)}}}

Since we are given AP=PQ=QR

{{{RB=AP*sin(alpha)+AP*sin(2alpha)+AP*sin(3alpha)}}}

{{{RB=AP(sin(alpha)+sin(2alpha)+sin(3alpha))}}}
---------------------------------------

Exactly similar to above using cosines instead of sines and

AS, PU and QV instead of PSQU and RV, you can prove

{{{RB=AP(sin(alpha)+sin(2alpha)+sin(3alpha))}}}

then 

{{{tan(BAR)=RB/AB = (AP(sin(alpha)+sin(2alpha)+sin(3alpha)))/

(AP(sin(alpha)+sin(2alpha)+sin(3alpha)))}}}

Cancel the AP's and get

tan(BAR) = {{{(sin(alpha)+sin(2alpha)+sin(3alpha))/(cos(alpha)+cos(2alpha)+cos(3alpha))}}}


  Edwin</pre>