Answer:
[tex]x=\frac{10\sqrt{6}}{7}\ in[/tex]
Step-by-step explanation:
step 1
In the right triangle DOC
Find the measure of side DO
Applying the Pythagoras Theorem
[tex]DC^{2}=DO^{2}+OC^{2}[/tex]
substitute the given values
[tex]7^{2}=DO^{2}+5^{2}[/tex]
[tex]DO^{2}=7^{2}-5^{2}[/tex]
[tex]DO^{2}=49-25[/tex]
[tex]DO^{2}=24[/tex]
[tex]DO=2\sqrt{6}\ in[/tex]
step 2
In the right triangle DOC
Find the sine of angle ∠ODC
sin(∠ODC)=OC/DC
substitute
[tex]sin(ODC)=5/7[/tex] -----> equation A
step 3
In the right triangle DOP
Find the sine of angle ∠ODP
sin(∠ODP)=OP/DO
substitute
[tex]sin(ODP)=x/2\sqrt{6}[/tex] -----> equation B
step 4
Find the value of x
In this problem
∠ODC=∠ODP
so
equate equation A and equation B
[tex]5/7=x/2\sqrt{6}[/tex]
[tex]x=\frac{10\sqrt{6}}{7}\ in[/tex]
