The problem asks to show that the size of the angle $FGH$ is $65^\circ$, correct to the nearest degree. We are provided with previous calculations where $IGF1 = \sqrt{10.96} = 3.31$ (approximately).

GeometryAnglesTrianglesCosine RuleGeometric Proofs
2025/5/27

1. Problem Description

The problem asks to show that the size of the angle FGHFGH is 6565^\circ, correct to the nearest degree. We are provided with previous calculations where IGF1=10.96=3.31IGF1 = \sqrt{10.96} = 3.31 (approximately).

2. Solution Steps

Without additional context or a diagram, it is impossible to definitively show that the angle FGHFGH is 6565^\circ. I need more information on the geometric relationship between the points F,G,H,F, G, H, and any other points or lengths given in the problem. The previous calculation IGF1=10.963.31IGF1 = \sqrt{10.96} \approx 3.31 suggests that this value could be a side length, area, or other parameter that can be used to determine FGH\angle FGH.
However, let's consider an example scenario using the Cosine Rule if we had the side lengths of the triangle FGHFGH. Let FG=aFG=a, GH=bGH=b, and FH=cFH=c. Then the Cosine Rule states:
c2=a2+b22abcos(G)c^2 = a^2 + b^2 - 2ab \cos(G)
Therefore,
cos(G)=a2+b2c22ab\cos(G) = \frac{a^2 + b^2 - c^2}{2ab}
If we are given the lengths of the three sides of the triangle, then we can substitute in the known values to calculate cos(G)\cos(G) and then use the inverse cosine function (cos1\cos^{-1}) to find the angle G=FGHG = \angle FGH.
Without a specific diagram and the lengths of relevant sides, I cannot proceed.
Let me assume that we are given FG=3FG=3, GH=4GH=4, and FH=5.3FH=5.3. In this case, we would have:
cos(FGH)=32+425.322(3)(4)=9+1628.0924=3.0924=0.12875\cos(\angle FGH) = \frac{3^2 + 4^2 - 5.3^2}{2(3)(4)} = \frac{9 + 16 - 28.09}{24} = \frac{-3.09}{24} = -0.12875
FGH=cos1(0.12875)97.4\angle FGH = \cos^{-1}(-0.12875) \approx 97.4^\circ. This is incorrect.
Let's assume we are given FG=4.5FG=4.5, GH=3.2GH=3.2, and FH=4FH=4.
cos(FGH)=4.52+3.22422(4.5)(3.2)=20.25+10.241628.8=14.4928.80.503125\cos(\angle FGH) = \frac{4.5^2 + 3.2^2 - 4^2}{2(4.5)(3.2)} = \frac{20.25 + 10.24 - 16}{28.8} = \frac{14.49}{28.8} \approx 0.503125
FGH=cos1(0.503125)59.8\angle FGH = \cos^{-1}(0.503125) \approx 59.8^\circ (close to 6060^\circ).
Again, I cannot show that the angle is exactly 6565^\circ without more information.

3. Final Answer

Insufficient information to solve the problem. More context or a diagram is needed.

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