On a morning of a day when the sun will pass directly overhead, the shadow of a 60-ft building on level ground is 25 ft long. at the moment in question, the angle theta the sun makes with the ground is increasing at the rate of 0.24 degrees/min. at what rate is the shadow decreasing? remember to use radians in your calculations. express your answer in inches per minute.

The shadow is decreasing at the rate of 3.55 inch/min

Step-by-step explanation:

The height of the building = 60ft

The shadow of the building on the level ground is 25ft long

Ѳ is increasing at the rate of 0.24°/min

Using SOHCAHTOA,

Tan Ѳ = opposite/ adjacent

= height of the building / length of the shadow

Tan Ѳ = h/x

X= h/tan Ѳ

Recall that tan Ѳ = sin Ѳ/cos Ѳ

X= h/x (sin Ѳ/cos Ѳ)

Differentiate with respect to t

dx/dt = (-h/sin²Ѳ)dѲ/dt

When x= 25ft

tanѲ = h/x

= 60/25

Ѳ= tan^-1(60/25)

= 67.38°

dѲ/dt= 0.24°/min

Convert the height in ft to inches

1 ft = 12 inches

Therefore, 60ft = 60*12

= 720 inches

Convert degree/min to radian/min

1°= 0.0175radian

Therefore, 0.24° = 0.24 * 0.0175

= 0.0042 radian/min

Recall that

dx/dt = (-h/sin²Ѳ)dѲ/dt

= (-720/sin²(67.38))*0.0042

= (-720/0.8521)*0.0042

-3.55 inch/min

Therefore, the rate at which the length of the shadow of the building decreases is 3.55 inches/min

since no one answered yet i will give it a shot, not completely sure tho.

i think it should be y=cos(x+2pi), (third answer, answer c)

good luck.

i believe that the answer is y+6=1/3(x+2)