This doesn't look like a rotation to me. I graphed it on desmos and somehow see that the triangle has shrinked. I think that point F should be stretched more for the two triangles to be equal.

Question

175. Describe a transformation that carries the triangle with vertices (0, 0), (13, 0), and (3, 2) onto the triangle with vertices (0,0), (12,5), and (2,3). Where does your transformation send the point (6, 0)?

Answer 1

draw a diagram on graph paper (draw the triangles)

use matrices

point (3,2) goes to point (2,3)

2 cosθ -sinθ 3

=

3 sinθ cosθ 2

2=3cosθ-2sinθ

3=3sinθ+2cosθ

rewrite the second equation

2=3cosθ-2sinθ

3=2cosθ+3sinθ

multiply all terms in top equation by 3 and multiply all terms in bottom equation by 2 to get

6=9cosθ-6sinθ

6=4cosθ+6sinθ

add the equations to eliminate sinθ

12=13cosθ

cosθ=12/13

use cos^-1θ=12/13 (0.9230769) on your scientific calculator

θ=22.61986495° (round as needed)

the transformation is a 22.6198º counterclockwise rotation about the origin

you can check this out using the point (13,0) which goes to the point (12,5) using the same matrix formula

use the same matrix formula for the point (6,0)

you get...

x=6cos(22.6198º) and y=6sin(22.6198º)

x=5.53846 and y=2.30768

the transformation takes the point (6,0) to the point ((5.53846,2.30768)

Answer 2

check out this, 13²=169=12²+5²

The rotation is around the origin point.

The angle is determined by trigonometry.

x'=xcosB+ysinB

y'=(-)xsinB+ycosB

for point 13,0 to point 12,5

correction, had x',y' and x,y switched

13=12cosB+5sinB

0=(-)12sinB+5cosB

from second equation, B=+.4radians

for 6,0

6=x*12/13+y*5/13

0=-x*5/13+y*12/13

y=2.3

x=5.52