Question

question
in $\triangle bcd$, $\overline{bd}$ is extended through point $d$ to point $e$, $m\angle bcd = (3x + 14)\degree$, $m\angle cde = (9x - 17)\degree$, and $m\angle dbc = (3x + 20)\degree$. find $m\angle dbc$.

Explanation:

Step1: Apply the exterior - angle theorem

The exterior - angle theorem states that the measure of an exterior angle of a triangle is equal to the sum of the measures of the two non - adjacent interior angles.
In \(\triangle BCD\), \(\angle CDE\) is an exterior angle. So, \(m\angle CDE=m\angle BCD + m\angle DBC\).
Substitute the given angle expressions: \((9x - 17)=(3x + 14)+(3x + 20)\).

Step2: Solve the equation for \(x\)

Simplify the right - hand side of the equation: \((3x + 14)+(3x + 20)=3x+3x + 14 + 20=6x+34\).
The equation becomes \(9x - 17=6x + 34\).
Subtract \(6x\) from both sides: \(9x-6x - 17=6x-6x + 34\), which gives \(3x-17 = 34\).
Add \(17\) to both sides: \(3x-17 + 17=34 + 17\), so \(3x=51\).
Divide both sides by \(3\): \(x=\frac{51}{3}=17\).

Step3: Find \(m\angle DBC\)

Substitute \(x = 17\) into the expression for \(m\angle DBC\).
\(m\angle DBC=(3x + 20)^{\circ}\).
Replace \(x\) with \(17\): \(m\angle DBC=(3\times17 + 20)^{\circ}=(51+20)^{\circ}=71^{\circ}\).

Answer:

\(71^{\circ}\)