Question

1. find $mangle bdc$.

(a.) $25^{circ}$
(b.) $52^{circ}$
(c.) $2^{circ}$
(d.) $38^{circ}$

Explanation:

Step1: Set up the equation

Since \(\angle BDC+(6x - 28)^{\circ}+(9x - 47)^{\circ}=90^{\circ}\) (right - angle property).
Combine like terms: \(\angle BDC+(15x-75)^{\circ}=90^{\circ}\).
But also, from the figure (assuming \(\angle BDC = 28^{\circ}\) is wrong initial thought, actually using the fact that \(6x-28 = 9x - 47\) is wrong. Correct way: since \((6x - 28)+(9x - 47)=90\) (sum of two non - \(\angle BDC\) angles in the right - angle).
\(6x+9x-28 - 47=90\).
\(15x-75 = 90\).

Step2: Solve for \(x\)

Add \(75\) to both sides: \(15x=90 + 75\).
\(15x=165\).
Divide both sides by \(15\): \(x = 11\).

Step3: Find \(\angle BDC\)

Substitute \(x = 11\) into \((6x - 28)\) (assuming \(\angle BDC=(6x - 28)\) from the figure's structure).
\(\angle BDC=6\times11-28\).
\(\angle BDC=66 - 28\).
\(\angle BDC = 38^{\circ}\) (wait, no, re - check. Wait, actually, if we assume the three angles \((6x - 28)\), \(\angle BDC\), \((9x - 47)\) form a right - angle. But if we use the property that \((6x - 28)+(9x - 47)=90\) (sum of two acute angles in a right - triangle). Then \(x = 11\). Then \(\angle BDC\): if we assume \(\angle BDC=(6x - 28)\), wrong. Wait, no, looking at the figure (assuming standard angle - addition in a right - angle). Let's re - do.

Since \((6x - 28)+(9x - 47)=90\) (sum of two non - \(\angle BDC\) angles in a right - angle).
\(15x-75 = 90\), \(x = 11\).
If \(\angle BDC=(6x - 28)\), then \(6\times11-28=38\) (no, wait, no. Wait, actually, if we assume \(\angle BDC\) is one of the angles. Wait, another approach:

We know that \((6x - 28)+(9x - 47)=90\) (complementary angles).
\(15x=90 + 75\), \(x = 11\).
Now, if \(\angle BDC=(6x - 28)\), \(6\times11-28=38\) (no, wait, no. Wait, hold on, the problem might have a typo. Wait, another way: assume \(\angle BDC\) is calculated as follows.

Since \((6x - 28)+(9x - 47)=90\) (sum of two angles adjacent to \(\angle BDC\) forming a right - angle).
\(15x=165\), \(x = 11\).
If \(\angle BDC=(9x - 47)\), \(9\times11-47=99 - 47 = 52\) (no). Wait, no, wait the figure: if \(BD\perp DA\), then \(\angle BDA = 90^{\circ}\). Assume \(\angle BDC+(6x - 28)+(9x - 47)=90\) (no, no, that's not. Wait, no, the standard is \((6x - 28)+(9x - 47)=90\) (sum of two angles in a right - angle). Then \(x = 11\). Then if \(\angle BDC\) is calculated as: if we assume the options, check \(x = 11\) in each.

Option a: \(25^{\circ}\): if \(6x - 28=25\), \(6x=53\), \(x=\frac{53}{6}\approx8.83\); if \(9x - 47=25\), \(9x=72\), \(x = 8\).

Option b: \(52^{\circ}\): if \(9x - 47=52\), \(9x=99\), \(x = 11\).

Answer:

b. \(52^{\circ}\)