Question 1177016
{{{drawing(400,400, -1.2,1.2,-1.2,1.2, circle(0,0,1),
line(cos(40*pi/180),sin(40*pi/180),cos(40*pi/180),sin(40*pi/180)),
line(cos(40*pi/180),sin(40*pi/180),cos(290*pi/180),sin(290*pi/180)),
line(cos(290*pi/180),sin(290*pi/180),cos(240*pi/180),sin(240*pi/180)),
line(cos(240*pi/180),sin(240*pi/180),cos(160*pi/180),sin(160*pi/180)),
line(cos(160*pi/180),sin(160*pi/180),cos(40*pi/180),sin(40*pi/180)),

locate(.8,.7,E), locate(.35,-.96,L), locate(-1.02,.38,V), locate(-.5,-.9,O),

red(line(cos(160*pi/180),sin(160*pi/180),cos(290*pi/180),sin(290*pi/180)),
locate(.03,.08,K),line(0,0,cos(160*pi/180),sin(160*pi/180)),line(0,0,cos(290*pi/180),sin(290*pi/180)),locate(.6,.65,55^o),locate(-.21,.03,110^o),locate(-.47,.27,25),locate(.2,-.4,25),locate(-.5,-.7,125^o)))  )}}} 
<pre>
Let the center be K. Draw radii KV and KL [which are 25 in. each. We'll get
that in part d)].  Let's answer part b) first
</pre>b)Angle measure of Minor Arc VL<pre>
Since inscribed angle VEL is 55°, then 55° is one-half the measure of
its intercepted arc.  So minor arc VL is twice 55°, or 110°.

Let's do part a) next
</pre>a)Length of Major Arc VL<pre>
The measure of major arc VL = 360°-the measure of minor arc VL 
= 360-°110°=250° 

Let's do part c) next
</pre>c)Angle VOL<pre>
Since inscribed angle VOL subtends major acc VL, which is 250°, then 250° is
one-half the measure of its intercepted arc. So angle VOL is 125°  
</pre>d)Length of Chord VL<pre>
Since we found minor arc VL to have measure 110° in part a), Central angle
VKL also has measure 110°.

Now we will draw KP perpendicular to VL.  

{{{drawing(400,400, -1.2,1.2,-1.2,1.2, circle(0,0,1),
line(cos(40*pi/180),sin(40*pi/180),cos(40*pi/180),sin(40*pi/180)),
line(cos(40*pi/180),sin(40*pi/180),cos(290*pi/180),sin(290*pi/180)),
line(cos(290*pi/180),sin(290*pi/180),cos(240*pi/180),sin(240*pi/180)),
line(cos(240*pi/180),sin(240*pi/180),cos(160*pi/180),sin(160*pi/180)),
line(cos(160*pi/180),sin(160*pi/180),cos(40*pi/180),sin(40*pi/180)),
green(line(0,0,-.2988362387,-.2988362387),locate(-.21,.03,55^o),
locate(-.34,-.31,P)
),
locate(.8,.7,E), locate(.35,-.96,L), locate(-1.02,.38,V), locate(-.5,-.9,O),

red(line(cos(160*pi/180),sin(160*pi/180),cos(290*pi/180),sin(290*pi/180)),
locate(.03,.08,K),line(0,0,cos(160*pi/180),sin(160*pi/180)),line(0,0,cos(290*pi/180),sin(290*pi/180)),locate(.6,.65,55^o),locate(-.47,.27,25),locate(.2,-.4,25),locate(-.5,-.7,125^o)))  )}}} 

</pre>d)Length of Chord VL<pre>
Since triangle VKL is isosceles, KP bisects angle VKL, which is 110°, and
angle VKP is 55°.

{{{sin(VKP)=opposite/hypotenuse=VP/VK}}}
{{{sin(55^o)=VP/25}}}
{{{25sin(55^o)=VP}}}
{{{20.47880111=VP}}}

Since chord VL is twice VP, chord VL has length 40.95760221,

Edwin</pre>