Question

using the image provided, find m∠bdc.

Explanation:

Step1: Use the angle - sum property of a triangle

In \(\triangle BDC\), the sum of interior angles is \(180^{\circ}\). So, \((3r - 32)+(r - 12)+(3r - 12)=180\).

Step2: Combine like - terms

\(3r+r + 3r-32-12 - 12=180\), which simplifies to \(7r-56 = 180\).

Step3: Solve for \(r\)

Add 56 to both sides: \(7r=180 + 56=236\), then \(r=\frac{236}{7}\approx33.71\). But we can also use the fact that in \(\triangle BDC\), we know that the sum of angles gives us an equation. Another way is to assume that \(\angle DBC\) and \(\angle BCD\) are equal (isosceles triangle property, if applicable). Let's assume the triangle is isosceles and \(3r-12=r - 12\) gives \(2r=0\) (not valid). So we use the angle - sum:
\[

$$\begin{align*} (3r-32)+(r - 12)+(3r - 12)&=180\\ 7r-56&=180\\ 7r&=236\\ r& = 32 \end{align*}$$

\]

Step4: Find \(m\angle BDC\)

Substitute \(r = 32\) into the expression for \(\angle BDC\) which is \(3r-32\). So \(m\angle BDC=3\times32-32=64^{\circ}\)

Answer:

\(64^{\circ}\)