Answer:
45.1feet
Step-by-step 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
M = 90 kg,
On Mars: g = 3.71 m/s²
Weight = m · g = 90 kg · 3.71 m/s² = 333.9 N
1 N = 0.2248 lbf
333.9 · 0.2248 = 75.06072 lbf ≈ 75 lbf
Step-by-step explanation:
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