Triangle FGH has vertices F(0, 10), G(10, 0), and HR(1, -1). After rotating triangle FGH counterclockwise about the origin 75º, the coordinates of G' to the nearest hundredth are (2.59, ?).

I will assume that HR(1,-1) is a typo and you meant

H(1,-1). Doesn't matter anyway, since it does not enter the calculation.
The question reduces to a simple problem
Rotae OG counterclockwise to OG' through and angle of 75°
Clearly OG' has to be 10 , since OG = 10

We could just use Pythagoras,
2.59^2 + y^2 = 10^2
y^2 = 93.2919
y = 9.66

notice that tanØ = 9.66/2.59 ---> 74.989° or 75°

the given information of G'( 2.59, ??) was actually redundant information and we could have found G' by simply using trig

x/10 = cos 75 ---> x = 10cos75 = 2.588
y/10 = sin 75 ---> y = 10sin75 = 9.66

To find the coordinates of G' after rotating triangle FGH counterclockwise about the origin, we can use rotation formulas.

The rotation formula for counterclockwise rotation about the origin is:

x' = x * cos(theta) - y * sin(theta)
y' = x * sin(theta) + y * cos(theta)

where (x', y') are the new coordinates after rotation, (x, y) are the original coordinates, and theta is the angle of rotation.

In this case, the original coordinates of G are (10, 0), and we want to rotate it counterclockwise about the origin by 75º.

Substituting these values into the rotation formulas, we have:

x' = 10 * cos(75º) - 0 * sin(75º)
y' = 10 * sin(75º) + 0 * cos(75º)

Using a calculator or a trigonometric table, we find that cos(75º) is approximately 0.2588 and sin(75º) is approximately 0.9659.

Now, we can calculate the new coordinates:

x' = 10 * 0.2588 - 0 * 0.9659 = 2.588
y' = 10 * 0.9659 + 0 * 0.2588 = 9.659

Rounding these values to the nearest hundredth, G' has coordinates (2.59, 9.66).