Question

in isosceles trapezoid abcd below, $angle abc$ measures $2(x - 10)$ , and $angle adc$ measures $3x$.
find the value of $x$
$x = $
find the measures of $angle dab$ and $angle abc$
the $mangle dab = $

Explanation:

Step1: Recall isosceles trapezoid property

In an isosceles trapezoid, consecutive angles between the bases are supplementary (sum to $180^\circ$). Here, $\angle ABC$ and $\angle BCD$ are consecutive angles, so:
$$2(x-10) + 3x = 180$$

Step2: Simplify and solve for $x$

Expand and combine like terms:
$$2x - 20 + 3x = 180$$
$$5x - 20 = 180$$
Add 20 to both sides:
$$5x = 200$$
Divide by 5:
$$x = 40$$

Step3: Calculate $\angle ABC$

Substitute $x=40$ into $\angle ABC = 2(x-10)$:
$$\angle ABC = 2(40-10) = 2(30) = 60^\circ$$

Step4: Find $\angle DAB$

In an isosceles trapezoid, base angles are equal. $\angle DAB = \angle ABC$, so:
$$\angle DAB = 60^\circ$$

Answer:

$x = 40$
$m\angle DAB = 60^\circ$
$m\angle ABC = 60^\circ$