In the given figure, we have two right angled triangles:
1) Triangle ABC
2) Triangle CDB
Using pythagorean theorem, we can write equations for both triangles.
For triangle ABC:
[tex] (15+x)^{2}= 17^{2}+ y^{2} [/tex]
For triangle CDB:
[tex] y^{2}= 8^{2}+ x^{2} [/tex]
Using the value of y² in the first equation, we get:
[tex] (15+x)^{2}= 289 + 64 + x^{2} \\ \\
225+30x+ x^{2} =353 + x^{2} \\ \\
30x=128 \\ \\
x= \frac{128}{30} \\ \\
x= \frac{64}{15}
[/tex]
[tex]y^{2}= 8^{2}+ x^{2} \\ \\
y^{2}=64+ ( \frac{64}{15} )^{2} \\ \\
y^{2}= \frac{18496}{25} \\ \\
y= \frac{136}{15} [/tex]
Thus the d option gives the correct values of x and y
