Question 187025
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We'll do the 1st triangle ABC :
Let's see the triangle:
{{{drawing(300,300,-3,3,-1,14,triangle(-2,0,0,12,2,0),green(line(0,0,0,12)),locate(0,12.8,A),locate(-2.3,0,B),locate(2.3,0,C),red(locate(0,-.3,9)),red(locate(-1.8,1.2,72^o)),red(locate(-1.5,6,13)),green(locate(0.2,6,D)),green(line(-.25,0,-.25,.5)),green(line(-.25,.5,.25,.5)),green(line(.25,.5,.25,0)),green(locate(.3,1,E)))}}}
As shown: we need AC=?; A=?; C?
We need to get 1st Length "D":
{{{sin72^o=D/13}}}--->{{{D=(sin72^o)(13)=12.36}}}
By Pyth Theorem, we get:
{{{13^2=D^2+(BE)^2 }}}---->{{{169=12.36^2+(BE)^2}}}
{{{(BE)^2=169-152.77=16.23}}}
{{{BE=sqrt(16.23)=4.03}}}
Then, {{{EC=9-BE=9-4.03=4.97}}}
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Now, for length "AC": --->{{{AC^2=D^2+EC^2=12.36^2+4.97^2}}}
{{{AC=sqrt(177.47)}}}----><font color="red">AC=13.32</font>
For Angle C:----> {{{sin(C)=opp/hyp=D/AC=12.36/13.32=0.9279279}}}
{{{(C)=sin^-1(0.9279279)}}}----> <font color="red">C=68.11^o</font>
For Angle A: {{{A+B+C=180^0}}}
{{{A+72+68.11=180}}}----->{{{A=180-72-68.11}}}---><font color="red">A=39.89^o</font>: See below,
{{{drawing(300,300,-3,3,-1,14,triangle(-2,0,0,12,2,0),locate(0,12.8,A),locate(-2.3,0,B),locate(2.3,0,C),red(locate(0,-.3,9)),red(locate(-1.8,1.2,72^o)),red(locate(-1.5,6,13)),blue(locate(-.3,10.5,39.89^o)),blue(locate(1.2,1.2,68.11^o)),blue(locate(1.3,6,13.32)))}}}
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Thank you,
Jojo</pre>