Answer:
45.1feet
Explanation:
Given the following
∠I=90°
∠G=62°, and
GH = 96 feet = Hypotenuse
Required
IG = Adjacent side
Using the SOH CAH TOA identity
Cos theta = Adj/hyp
Cos 62 =IG/96
IG = 96cos62
IG = 96(0.4695)
IG = 45.1feet
Hence the length of IG to the nearest tenth is 45.1feet