Jay, Kay, Ray found themselves far apart for lunch while working in a field. Jay could see Kay, then turned 75 degrees and see Ray. Kay could see Ray, then turn 51 degrees and see Jay. Ray could see jay, then turn 54 degrees and see Kay. Which two were the furthest apart?

A) Ray and Jay
B) Kay and Ray
C) jay and Kay
D) Kay and Ray were the same distance as Ray and Jay

I believe the answer is A but can someone please check this for me and if I made a mistake can someone please explain to me why I got it wrong

Sorry I meant to put geometry as the subject

I made a rough sketch of triangle JKR and filled in the angles.

A basic property of triangles is that in any triangle,
the smallest side is opposite the smallest angle, and the largest side is opposite the largest angle

Since angle J of 75° is the largest, side RK must be the largest
so it looks like B)

Okay thank you so much, the question really confused me but I understand what you mean thank you so much

To determine which two individuals were the furthest apart, we need to compare the angles they turned.

From the given information:
- Jay turned 75 degrees and saw Kay.
- Kay turned 51 degrees and saw Ray.
- Ray turned 54 degrees and saw Jay.

To visualize this, we can draw a diagram:
```
R (Ray)
/
/
J (Jay)
\
\
K (Kay)
```

Now, let's calculate the angles between each pair of individuals to determine the answer.

1. Angle between Jay and Kay:
Jay turned 75 degrees, so the angle between Jay and Kay is 180 - 75 = 105 degrees.

2. Angle between Kay and Ray:
Kay turned 51 degrees, and Ray turned 54 degrees. Thus, the angle between Kay and Ray is 51 + 54 = 105 degrees.

3. Angle between Ray and Jay:
Ray turned 54 degrees, so the angle between Ray and Jay is 180 - 54 = 126 degrees.

Based on the calculations, the two individuals who were the furthest apart are Ray and Jay. Therefore, the correct answer is A) Ray and Jay.