Answer:
[tex]BC=4\sqrt{5}\ units[/tex]
Step-by-step explanation:
see the attached figure with letters to better understand the problem
step 1
In the right triangle ACD
Find the length side AC
Applying the Pythagorean Theorem
[tex]AC^2=AD^2+DC^2[/tex]
substitute the given values
[tex]AC^2=16^2+8^2[/tex]
[tex]AC^2=320[/tex]
[tex]AC=\sqrt{320}\ units[/tex]
simplify
[tex]AC=8\sqrt{5}\ units[/tex]
step 2
In the right triangle ACD
Find the cosine of angle CAD
[tex]cos(\angle CAD)=\frac{AD}{AC}[/tex]
substitute the given values
[tex]cos(\angle CAD)=\frac{16}{8\sqrt{5}}[/tex]
[tex]cos(\angle CAD)=\frac{2}{\sqrt{5}}[/tex] ----> equation A
step 3
In the right triangle ABC
Find the cosine of angle BAC
[tex]cos(\angle BAC)=\frac{AC}{AB}[/tex]
substitute the given values
[tex]cos(\angle BAC)=\frac{8\sqrt{5}}{16+x}[/tex] ----> equation B
step 4
Find the value of x
In this problem
[tex]\angle CAD=\angle BAC[/tex] ----> is the same angle
so
equate equation A and equation B
[tex]\frac{8\sqrt{5}}{16+x}=\frac{2}{\sqrt{5}}[/tex]
solve for x
Multiply in cross
[tex](8\sqrt{5})(\sqrt{5})=(16+x)(2)\\\\40=32+2x\\\\2x=40-32\\\\2x=8\\\\x=4\ units[/tex]
[tex]DB=4\ units[/tex]
step 5
Find the length of BC
In the right triangle BCD
Applying the Pythagorean Theorem
[tex]BC^2=DC^2+DB^2[/tex]
substitute the given values
[tex]BC^2=8^2+4^2[/tex]
[tex]BC^2=80[/tex]
[tex]BC=\sqrt{80}\ units[/tex]
simplify
[tex]BC=4\sqrt{5}\ units[/tex]
