Question

challenge given that m∠a=(18x)°, m∠c=(8x + 23)°, and m∠d=113°, what is m∠b? m∠b=□°

Explanation:

Step1: Recall the sum of angles in a quadrilateral

The sum of the interior angles of a quadrilateral is $360^{\circ}$. So, $m\angle A + m\angle B+m\angle C + m\angle D=360^{\circ}$.

Step2: Substitute the given angle - measures

We know that $m\angle A=(18x)^{\circ}$, $m\angle C=(8x + 23)^{\circ}$, and $m\angle D = 113^{\circ}$. Substituting these into the equation from Step 1 gives us $(18x)+m\angle B+(8x + 23)+113 = 360$.

Step3: Combine like - terms

Combine the $x$ terms and the constant terms: $18x+8x+m\angle B+23 + 113=360$, which simplifies to $26x+m\angle B+136 = 360$.

Step4: Isolate the terms with $x$ and $\angle B$

Subtract 136 from both sides of the equation: $26x+m\angle B=360 - 136$, so $26x+m\angle B=224$.
We also know that for a quadrilateral, we can assume it is a cyclic quadrilateral (if no other information is given, and since we need to solve for $x$ to find the angle). In a cyclic quadrilateral, the sum of opposite angles is $180^{\circ}$. So, $m\angle A+m\angle C = 180^{\circ}$ (assuming it is cyclic, a common approach when dealing with angle - finding in quadrilaterals with limited information).
Set up the equation $18x+8x + 23=180$.

Step5: Solve for $x$ in the new equation

Combine like - terms: $26x+23 = 180$. Subtract 23 from both sides: $26x=180 - 23=157$. Then $x=\frac{157}{26}\approx6.04$. But if we assume the quadrilateral is cyclic and use the property of opposite angles:
$18x+8x+23 = 180$.
$26x=180 - 23=157$.
$x=\frac{157}{26}$.
Now, using the sum of angles of a quadrilateral $m\angle A + m\angle B+m\angle C + m\angle D=360^{\circ}$.
$18x+m\angle B+(8x + 23)+113 = 360$.
Substitute $x$ values:
$18\times\frac{157}{26}+m\angle B+(8\times\frac{157}{26}+23)+113 = 360$.
Another way, using the property of cyclic quadrilateral $m\angle B + m\angle D=180^{\circ}$ (opposite angles of a cyclic quadrilateral).
Since $m\angle D = 113^{\circ}$, then $m\angle B=180 - 113$.

Answer:

$67^{\circ}$