Question

in rhombus $mjkl$; $\angle jbk = 10(x + 1)^2$, $jk = 15$, $jb = 9$, $bk = y$. solve for $x$ and $y$.

Explanation:

Step1: Use the property of rhombus diagonals

In a rhombus, the diagonals are perpendicular to each other. So, $\angle JBK = 90^{\circ}$.
Set up the equation $10(x + 1)^{2}=90$.
Divide both sides by 10: $(x + 1)^{2}=\frac{90}{10}=9$.
Take the square root of both sides: $x + 1=\pm3$.
Case 1: $x+1 = 3$, then $x=3 - 1=2$.
Case 2: $x + 1=-3$, then $x=-3 - 1=-4$. But since we are dealing with a geometric figure (length - related in terms of angle expression context, we take $x = 2$ (assuming non - negative in geometric sense for this type of problem setup).

Step2: Use the Pythagorean theorem

In right - triangle $JBK$, by the Pythagorean theorem $JB^{2}+BK^{2}=JK^{2}$.
We know $JK = 15$ and $JB = 9$. Substitute into the formula: $9^{2}+y^{2}=15^{2}$.
$81+y^{2}=225$.
Subtract 81 from both sides: $y^{2}=225 - 81=144$.
Take the square root of both sides: $y=\sqrt{144}=12$ (we take the positive value since $y$ represents a length).

Answer:

$x = 2$; $y = 12$