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2. m∠abe=(4x + 2)° and m∠abd=(5x + 16)°. find x, m∠abd and m∠cbd.

Question

  1. m∠abe=(4x + 2)° and m∠abd=(5x + 16)°. find x, m∠abd and m∠cbd.

Explanation:

Step1: Set up equation using vertical - angle property

Vertical angles are equal. $\angle ABE$ and $\angle CBD$ are vertical angles, and we assume $\angle ABD$ and $\angle EBC$ are vertical angles. Since $\angle ABE$ and $\angle ABD$ are supplementary (a linear - pair), we have $(4x + 2)+(5x + 16)=180$.
$4x+2 + 5x+16=180$
$9x+18 = 180$

Step2: Solve for x

Subtract 18 from both sides of the equation:
$9x=180 - 18$
$9x=162$
Divide both sides by 9:
$x=\frac{162}{9}=18$

Step3: Find m∠ABD

Substitute $x = 18$ into the expression for $\angle ABD$:
$m\angle ABD=(5x + 16)^{\circ}$
$m\angle ABD=5\times18+16$
$m\angle ABD = 90 + 16=106^{\circ}$

Step4: Find m∠CBD

Since $\angle ABE$ and $\angle CBD$ are vertical angles, $m\angle CBD=m\angle ABE$.
$m\angle ABE=(4x + 2)^{\circ}$, substituting $x = 18$ gives $m\angle ABE=4\times18+2=72 + 2=74^{\circ}$
So $m\angle CBD = 74^{\circ}$

Answer:

$x = 18$, $m\angle ABD=106^{\circ}$, $m\angle CBD = 74^{\circ}$