unit 4 triangle properties test
given:
• $ \angle a b d = 105 ^ { \circ } $
• $ \angle b d c = ( 4 x - 15 ) ^ { \circ } $
• $ \angle b c d = x ^ { \circ } $
find $ m \angle b d c $.

Question

Accepted Answer

# Explanation:
## Step1: Use the exterior - angle property of a triangle
The exterior - angle property of a triangle states that an exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. In $\triangle BCD$, $\angle ABD$ is an exterior angle. So, $\angle ABD=\angle BDC+\angle BCD$.
Given $\angle ABD = 105^{\circ}$, $\angle BDC=(4x - 15)^{\circ}$, and $\angle BCD=x^{\circ}$. Then $105=(4x - 15)+x$.
## Step2: Solve the equation for $x$
Simplify the right - hand side of the equation $105=(4x - 15)+x$:
$$
\begin{align*}
105&=4x- 15+x\
105&=5x-15
\end{align*}
$$
Add $15$ to both sides of the equation: $105 + 15=5x-15 + 15$, which gives $120 = 5x$.
Divide both sides by $5$: $x=\frac{120}{5}=24$.
## Step3: Find the measure of $\angle BDC$
Substitute $x = 24$ into the expression for $\angle BDC$. Since $\angle BDC=(4x - 15)^{\circ}$, then $\angle BDC=(4\times24-15)^{\circ}$.
$$
\begin{align*}
\angle BDC&=(96 - 15)^{\circ}\
&=81^{\circ}
\end{align*}
$$

# Answer:
$81^{\circ}$

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QUESTION IMAGE

Question

Explanation:

Step1: Use the exterior - angle property of a triangle

Step2: Solve the equation for \(x\)

Step3: Find the measure of \(\angle BDC\)

Answer: