Question

find m∠rsq and m∠tsq.
(15x - 43)°
(8x + 18)°
m∠rsq = (square^{circ})
m∠tsq = (square^{circ})

Explanation:

Step1: Set up equation

Since $\angle RST = 90^{\circ}$ and $\angle RST=\angle RSQ+\angle TSQ$, we have $(15x - 43)+(8x + 18)=90$.

Step2: Combine like - terms

Combining the $x$ terms and the constant terms on the left - hand side gives $15x+8x-43 + 18=90$, which simplifies to $23x-25 = 90$.

Step3: Solve for $x$

Add 25 to both sides of the equation: $23x=90 + 25=115$. Then divide both sides by 23, so $x=\frac{115}{23}=5$.

Step4: Find $m\angle RSQ$

Substitute $x = 5$ into the expression for $\angle RSQ$: $m\angle RSQ=15x-43=15\times5-43=75 - 43=32^{\circ}$.

Step5: Find $m\angle TSQ$

Substitute $x = 5$ into the expression for $\angle TSQ$: $m\angle TSQ=8x + 18=8\times5+18=40+18=58^{\circ}$.

Answer:

$m\angle RSQ = 32^{\circ}$
$m\angle TSQ = 58^{\circ}$