Question

for $\triangle abc$, $\angle a = 3x - 8$, $\angle b = 5x - 6$, and $\angle c = 4x + 2$. if $\triangle abc$ undergoes a dilation by a scale factor of $\frac{1}{2}$ to create $\triangle abc$ with $\angle a = 2x + 8$, $\angle b = 90 - x$, and $\angle c = 5x - 14$, which confirms that $\triangle abc \sim \triangle abc$ by the aa criterion?
a $\angle a = \angle a = 37^\circ$ and $\angle b = \angle b = 69^\circ$
b $\angle a = \angle a = 22^\circ$ and $\angle c = \angle c = 42^\circ$
c $\angle b = \angle b = 37^\circ$ and $\angle c = \angle c = 33^\circ$
d $\angle b = \angle b = 74^\circ$ and $\angle c = \angle c = 66^\circ$

Explanation:

Step1: Recall triangle angle sum

The sum of angles in a triangle is \(180^\circ\). So for \(\triangle ABC\), \(\angle A+\angle B+\angle C = 180^\circ\). Substitute the given expressions: \((3x - 8)+(5x - 6)+(4x + 2)=180\).
Simplify the left side: \(3x - 8+5x - 6+4x + 2 = 12x - 12\). So \(12x - 12 = 180\).
Add 12 to both sides: \(12x = 192\).
Divide by 12: \(x = 16\).

Step2: Calculate angles of \(\triangle ABC\)

• \(\angle A = 3x - 8 = 3(16)-8 = 48 - 8 = 40^\circ\) (Wait, maybe better to use AA criterion by equating angles from dilation. Since dilation preserves angle measures, \(\angle A=\angle A'\), \(\angle B=\angle B'\), \(\angle C=\angle C'\). Let's equate \(\angle B\) and \(\angle B'\): \(5x - 6 = 90 - x\).

Add \(x\) to both sides: \(6x - 6 = 90\).
Add 6: \(6x = 96\).
\(x = 16\) (same as before).

Now calculate \(\angle B = 5x - 6 = 5(16)-6 = 80 - 6 = 74^\circ\)
\(\angle C = 4x + 2 = 4(16)+2 = 64 + 2 = 66^\circ\)
Check \(\angle B' = 90 - x = 90 - 16 = 74^\circ\)
\(\angle C' = 5x - 14 = 5(16)-14 = 80 - 14 = 66^\circ\)
So \(\angle B=\angle B' = 74^\circ\) and \(\angle C=\angle C' = 66^\circ\), which satisfies AA criterion.

Answer:

D. \(\angle B = \angle B' = 74^\circ\) and \(\angle C = \angle C' = 66^\circ\)