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, ?).

Answer 1

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