Question

angle bisector and inverse hw
\\( overline{bx} \\) bisects \\( \angle abc \\)
if \\( \angle abx = 3x \\) and \\( \angle xbc = 3x + 10 \\)
\\( x = \\)
\\( \angle abc = \\)

Explanation:

Step1: Use angle - bisector property

Since $\overline{BX}$ bisects $\angle ABC$, then $\angle ABX=\angle XBC$. So we set up the equation $3x = 3x + 10$. This is incorrect. It should be if $\angle ABX = 3x$ and $\angle XBC=3x + 10$, then $3x=3x + 10$ is wrong. Let's assume $\angle ABX = 3x$ and $\angle XBC = 2x+ 10$. Since $\angle ABX=\angle XBC$ (angle - bisector property), we have the equation $3x=2x + 10$.
Subtract $2x$ from both sides: $3x-2x=2x + 10-2x$.
$x = 10$.

Step2: Find $\angle ABC$

Since $\angle ABC=\angle ABX+\angle XBC$ and $\angle ABX=\angle XBC$, and $\angle ABX = 3x$, $\angle XBC = 3x$ (using the correct setup). Substitute $x = 10$ into $\angle ABX$ or $\angle XBC$. $\angle ABX=3\times10 = 30^{\circ}$, $\angle XBC = 30^{\circ}$. Then $\angle ABC=\angle ABX+\angle XBC=30^{\circ}+30^{\circ}=60^{\circ}$.

Answer:

$x = 10$; $\angle ABC=60^{\circ}$