Step ![1](/tpl/images/0318/3274/8aa3f.png)
In the right triangle ADB
Find the length of the segment AB
Applying the Pythagorean Theorem
![AB^{2} =AD^{2}+BD^{2}](/tpl/images/0318/3274/d5645.png)
we have
![AD=5\ units\\BD=12\ units](/tpl/images/0318/3274/0f16e.png)
substitute the values
![AB^{2}=5^{2}+12^{2}](/tpl/images/0318/3274/60a64.png)
![AB^{2}=169](/tpl/images/0318/3274/949d6.png)
![AB=13\ units](/tpl/images/0318/3274/dc338.png)
Step ![2](/tpl/images/0318/3274/05229.png)
In the right triangle ADB
Find the cosine of the angle BAD
we know that
![cos(BAD)=\frac{adjacent\ side }{hypotenuse}=\frac{AD}{AB}=\frac{5}{13}](/tpl/images/0318/3274/b66bd.png)
Step ![3](/tpl/images/0318/3274/07736.png)
In the right triangle ABC
Find the length of the segment AC
we know that
![cos(BAC)=cos (BAD)=\frac{5}{13}](/tpl/images/0318/3274/05db5.png)
![cos(BAC)=\frac{adjacent\ side }{hypotenuse}=\frac{AB}{AC}](/tpl/images/0318/3274/d1976.png)
![\frac{5}{13}=\frac{AB}{AC}](/tpl/images/0318/3274/fbfa4.png)
![\frac{5}{13}=\frac{13}{AC}](/tpl/images/0318/3274/a8906.png)
solve for AC
![AC=(13*13)/5=33.8\ units](/tpl/images/0318/3274/a3562.png)
Step ![4](/tpl/images/0318/3274/ad743.png)
Find the length of the segment DC
we know that
![DC=AC-AD](/tpl/images/0318/3274/17591.png)
we have
![AC=33.8\ units](/tpl/images/0318/3274/30818.png)
![AD=5\ units](/tpl/images/0318/3274/29dfd.png)
substitute the values
![DC=33.8\ units-5\ units](/tpl/images/0318/3274/64d22.png)
![DC=28.8\ units](/tpl/images/0318/3274/354df.png)
Step ![5](/tpl/images/0318/3274/24fd4.png)
Find the length of the segment BC
In the right triangle BDC
Applying the Pythagorean Theorem
![BC^{2} =BD^{2}+DC^{2}](/tpl/images/0318/3274/40e6b.png)
we have
![BD=12\ units\\DC=28.8\ units](/tpl/images/0318/3274/b8a72.png)
substitute the values
![BC^{2}=12^{2}+28.8^{2}](/tpl/images/0318/3274/60ea7.png)
![BC^{2}=973.44](/tpl/images/0318/3274/9fbe1.png)
![BC=31.2\ units](/tpl/images/0318/3274/1fa17.png)
therefore
the answer is
![BC=31.2\ units](/tpl/images/0318/3274/1fa17.png)