Question 169639
Let's draw the pairs of Triangles to have a clearer view:
{{{drawing(400,400,-2,12,-2,15,triangle(10,0,0,0,0,14),red(line(4,0,4,8.4)),circle(4,8.4,.12),circle(4,0,.12),circle(10,0,.12),locate(7,-.25,6),locate(7,5,10),locate(-.85,7,14),locate(10.5,-.25,A),locate(-.85,0,C),locate(-.85,14,D),locate(4,9.3,E),locate(4,-.25,B),locate(4.3,1,90^o),locate(.3,1,90^o))}}}
.
We get 1st Angle {{{ALPHA}}}, which will be very helpful:
By Trigo Function: {{{Cos(alpha)=adj/hyp=AB/AE}}}
{{{alpha=cos^-1(AB/AE)=cos^-1(6/10)}}}
{{{alpha=53.13^o}}}
.
We solve for {{{BE}}}?, by Pyth.Teorem:
{{{AE^2=AB^2+BE^2}}}
{{{BE^2=AE^2-AB^2=10^2-6^2=100-36=64}}}
{{{BE=sqrt(64)}}} ---------->{{{highlight(BE=8)}}}
We solve for {{{AC}}}?, by Trigo function:
{{{tan(alpha)=opp/adj=DC/AC}}}
{{{AC=DC/tan(alpha)=14/tan53.13^o}}}
{{{highlight(AC=10.50)}}}
We solve for {{{AD}}}?, by Trigo function:
{{{sin(alpha)=opp/hyp=DC/AD}}}
{{{AD=DC/sin(alpha)=14/sin53.13^0}}}
{{{highlight(AD=17.50)}}}
See the 2 pairs of Triangle:
{{{drawing(400,400,-2,12,-2,15,triangle(10,0,0,0,0,14),red(line(4,0,4,8.4)),circle(4,8.4,.12),circle(4,0,.12),circle(10,0,.12),locate(7,-.25,6),locate(7,5,10),locate(-.85,7,14),locate(10.5,-.25,A),locate(-.85,0,C),locate(-.85,14,D),locate(4,9.3,E),locate(4,-.25,B),locate(4.3,1,90^o),locate(.3,1,90^o),locate(5,10,highlight(AD=17.50)),locate(2.2,5,highlight(BE=8)),locate(3.7,-1.2,highlight(AC=10.50)),locate(8,1,53.13^o))}}}
.
thank you,
Jojo