In trapezoid $ABCD$, $\overline{BC} \parallel \overline{AD}$, $\angle ABD = 105^\circ$, $\angle A = 43^\circ$, and $\angle C = 141^\circ$. Find $\angle CBD$, in degrees.
come on. the angles of any quadrilateral sum to 360
You do aops right? :)
lol when this time of year comes around...
Who even comes across these posts without looking for the answers? XD
does anyone know the answer lol
To find $\angle CBD$ in the given trapezoid, we can use the fact that the sum of the angles in a triangle is $180^\circ$.
Since $\overline{BC} \parallel \overline{AD}$, we can consider the triangles $\triangle CBD$ and $\triangle ABD$. The angle $\angle ABD = 105^\circ$ is the same as the angle $\angle BDC$ in triangle $\triangle CBD$ (corresponding angles when parallel lines are cut by a transversal).
Now, we can find the measure of angle $\angle CBD$ by subtracting the measures of the angles in triangle $\triangle CBD$ from $180^\circ$.
We have:
$\angle CBD = 180^\circ - \angle BDC - \angle BCD$
Since $\angle BDC = \angle ABD = 105^\circ$, and we don't know $\angle BCD$ yet, we can write the equation as:
$\angle CBD = 180^\circ - 105^\circ - \angle BCD$
We need to find $\angle BCD$. To do this, we can use the fact that the angles in a trapezoid add up to $360^\circ$.
Since $\angle A = 43^\circ$ and $\angle C = 141^\circ$, we know that $\angle BCD = 180^\circ - \angle A - \angle C$.
Substituting this value into the equation for $\angle CBD$, we have:
$\angle CBD = 180^\circ - 105^\circ - (180^\circ - \angle A - \angle C)$
Simplifying, we get:
$\angle CBD = 180^\circ - 105^\circ - 180^\circ + \angle A + \angle C$
Combining like terms, we have:
$\angle CBD = -105^\circ + \angle A + \angle C$
Finally, substituting the given values $\angle A = 43^\circ$ and $\angle C = 141^\circ$ into the equation, we can calculate:
$\angle CBD = -105^\circ + 43^\circ + 141^\circ$
Simplifying further, we get:
$\angle CBD = 79^\circ$
Therefore, $\angle CBD$ in the given trapezoid is $79^\circ$.