Step [tex]1[/tex]
In the right triangle ADB
Find the length of the segment AB
Applying the Pythagorean Theorem
[tex]AB^{2} =AD^{2}+BD^{2}[/tex]
we have
[tex]AD=5\ units\\BD=12\ units[/tex]
substitute the values
[tex]AB^{2}=5^{2}+12^{2}[/tex]
[tex]AB^{2}=169[/tex]
[tex]AB=13\ units[/tex]
Step [tex]2[/tex]
In the right triangle ADB
Find the cosine of the angle BAD
we know that
[tex]cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}[/tex]
Step [tex]3[/tex]
In the right triangle ABC
Find the length of the segment AC
we know that
[tex]cos(BAC)=cos (BAD)=\frac{5}{13}[/tex]
[tex]cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}[/tex]
[tex]\frac{5}{13}=\frac{AB}{AC}[/tex]
[tex]\frac{5}{13}=\frac{13}{AC}[/tex]
solve for AC
[tex]AC=(13*13)/5=33.8\ units[/tex]
Step [tex]4[/tex]
Find the length of the segment DC
we know that
[tex]DC=AC-AD[/tex]
we have
[tex]AC=33.8\ units[/tex]
[tex]AD=5\ units[/tex]
substitute the values
[tex]DC=33.8\ units-5\ units[/tex]
[tex]DC=28.8\ units[/tex]
Step [tex]5[/tex]
Find the length of the segment BC
In the right triangle BDC
Applying the Pythagorean Theorem
[tex]BC^{2} =BD^{2}+DC^{2}[/tex]
we have
[tex]BD=12\ units\\DC=28.8\ units[/tex]
substitute the values
[tex]BC^{2}=12^{2}+28.8^{2}[/tex]
[tex]BC^{2}=973.44[/tex]
[tex]BC=31.2\ units[/tex]
therefore
the answer is
[tex]BC=31.2\ units[/tex]
