Answer:
Part 1) [tex]z=4\sqrt{3}\ units[/tex]
Part 2) [tex]y=4\sqrt{2}\ units[/tex]
Part 3) [tex]x=4\sqrt{6}\ units[/tex]
Step-by-step explanation:
see the attached figure with letters to better understand the problem
step 1
Find the value of z
In the right triangle DBC
[tex]cos(C)=\frac{4}{z}[/tex] ----> equation A
In the right triangle ABC
[tex]cos(C)=\frac{z}{12}[/tex] ----> equation B
equate equation A and equation B
[tex]\frac{4}{z}=\frac{z}{12}[/tex]
[tex]z^{2}=48[/tex]
[tex]z=4\sqrt{3}\ units[/tex]
step 2
In the right triangle DBC
Applying the Pythagoras Theorem
[tex]z^{2}=y^{2}+4^{2}[/tex]
we have
[tex]z=4\sqrt{3}\ units[/tex]
substitute
[tex](4\sqrt{3})^{2}=y^{2}+4^{2}[/tex]
[tex]48=y^{2}+16[/tex]
[tex]y^{2}=48-16[/tex]
[tex]y^{2}=32[/tex]
[tex]y=4\sqrt{2}\ units[/tex]
step 3
Find the value of x
In the right triangle ABC
Applying the Pythagoras Theorem
[tex]12^{2}=x^{2}+z^{2}[/tex]
we have
[tex]z=4\sqrt{3}\ units[/tex]
substitute
[tex]12^{2}=x^{2}+(4\sqrt{3})^{2}[/tex]
[tex]144=x^{2}+48[/tex]
[tex]x^{2}=144-48[/tex]
[tex]x^{2}=96[/tex]
[tex]x=4\sqrt{6}\ units[/tex]
