if a person bends at the waist with a straight back making an angle of θ degrees with the horizontal, then…

if a person bends at the waist with a straight back making an angle of θ degrees with the horizontal, then the force f exerted on the back muscles can be modeled by the equation shown below, where w is the weight of the person. complete parts (a) through (c).\nf = \\frac{0.6w\\sin(\\theta + 90^{\\circ})}{\\sin 12^{\\circ}}\n(a) calculate f when w = 185 lb and θ = 20^{\\circ}.\nf = \\square lb\n(round to the nearest pound as needed.)

if a person bends at the waist with a straight back making an angle of θ degrees with the horizontal, then the force f exerted on the back muscles can be modeled by the equation shown below, where w is the weight of the person. complete parts (a) through (c).\nf = \\frac{0.6w\\sin(\\theta + 90^{\\circ})}{\\sin 12^{\\circ}}\n(a) calculate f when w = 185 lb and θ = 20^{\\circ}.\nf = \\square lb\n(round to the nearest pound as needed.)

Answer

Explanation:

Step1: Substitute the values of (W) and (\theta) into the formula

Given (W = 185) lb and (\theta=20^{\circ}), the formula is (F=\frac{0.6W\sin(\theta + 90^{\circ})}{\sin12^{\circ}}). First, calculate (\sin(\theta + 90^{\circ})). Using the trigonometric identity (\sin(A + B)=\sin A\cos B+\cos A\sin B), when (A=\theta = 20^{\circ}) and (B = 90^{\circ}), (\sin(20^{\circ}+90^{\circ})=\sin110^{\circ}=\sin(90^{\circ}+ 20^{\circ})=\cos20^{\circ}\approx0.9397).

Step2: Calculate the numerator

The numerator is (0.6\times W\times\sin(\theta + 90^{\circ})). Substitute (W = 185) and (\sin(\theta + 90^{\circ})\approx0.9397) into it. (0.6\times185\times0.9397=0.6\times173.8445 = 104.3067).

Step3: Calculate the denominator

The denominator is (\sin12^{\circ}\approx0.2079).

Step4: Calculate (F)

(F=\frac{104.3067}{0.2079}\approx502.7).

Answer:

(503) lb